ML Aggarwal Class 9 Maths Solutions Chapter 03 Expansions

 

Exercise 3.1

 

Question 1(i). By using standard formulae, expand the following:
\( (2x + 7y)^2 \)
Answer: \( (2x + 7y)^2 = (2x)^2 + 2(2x)(7y) + (7y)^2 = 4x^2 + 28xy + 49y^2 \)
In simple words: When you square a sum, the result is the first term squared, plus twice the product of both terms, plus the second term squared.

Exam Tip: Always apply the formula (a+b)² = a² + 2ab + b² carefully, making sure to square each term and multiply by 2 for the middle term.

 

Question 1(ii). By using standard formulae, expand the following:
\( \left( \frac{1}{2}x + \frac{2}{3}y \right)^2 \)
Answer: \( \left( \frac{1}{2}x + \frac{2}{3}y \right)^2 = \left( \frac{1}{2}x \right)^2 + 2\left( \frac{1}{2}x \right)\left( \frac{2}{3}y \right) + \left( \frac{2}{3}y \right)^2 = \frac{1}{4}x^2 + \frac{2}{3}xy + \frac{4}{9}y^2 \)
In simple words: Square each fraction separately, then add twice the product of the two terms together with the squared second term.

Exam Tip: Keep fractions in lowest terms and remember to multiply coefficients carefully when finding the middle term.

 

Question 2(i). By using standard formulae, expand the following:
\( \left( 3x + \frac{1}{2x} \right)^2 \)
Answer: \( \left( 3x + \frac{1}{2x} \right)^2 = (3x)^2 + 2(3x)\left(\frac{1}{2x}\right) + \left(\frac{1}{2x}\right)^2 = 9x^2 + 3 + \frac{1}{4x^2} \)
In simple words: Square the first term and the second term separately, then add twice their product in the middle.

Exam Tip: Notice that the middle term simplifies to 3, making the calculation easier - always simplify products before leaving them in final form.

 

Question 2(ii). By using standard formulae, expand the following:
\( (3x^2y + 5z)^2 \)
Answer: \( (3x^2y + 5z)^2 = (3x^2y)^2 + 2(3x^2y)(5z) + (5z)^2 = 9x^4y^2 + 30x^2yz + 25z^2 \)
In simple words: Square both terms and add their product multiplied by 2 to find the answer.

Exam Tip: When squaring a term like 3x²y, remember that the exponent applies to all parts - giving 9x⁴y².

 

Question 3(i). By using standard formulae, expand the following:
\( \left( 3x - \frac{1}{2x} \right)^2 \)
Answer: \( \left( 3x - \frac{1}{2x} \right)^2 = (3x)^2 - 2(3x)\left(\frac{1}{2x}\right) + \left(\frac{1}{2x}\right)^2 = 9x^2 - 3 + \frac{1}{4x^2} \)
In simple words: When you have a difference, square each part and subtract twice their product instead of adding.

Exam Tip: For (a-b)², remember the pattern is a² - 2ab + b², where the middle term is negative.

 

Question 3(ii). By using standard formulae, expand the following:
\( \left( \frac{1}{2}x - \frac{3}{2}y \right)^2 \)
Answer: \( \left( \frac{1}{2}x - \frac{3}{2}y \right)^2 = \left( \frac{1}{2}x \right)^2 - 2\left( \frac{1}{2}x \right)\left( \frac{3}{2}y \right) + \left( \frac{3}{2}y \right)^2 = \frac{1}{4}x^2 - \frac{3}{2}xy + \frac{9}{4}y^2 \)
In simple words: Use the difference formula by squaring both fractions and subtracting twice their product.

Exam Tip: Multiply fractions carefully - the middle term becomes \( -2 \times \frac{1}{2}x \times \frac{3}{2}y = -\frac{3}{2}xy \).

 

Question 4(i). By using standard formulae, expand the following:
\( (x+3)(x+5) \)
Answer: We know that \( (x+a)(x+b) = x^2 + (a+b)x + ab \).

\( \therefore (x+3)(x+5) = x^2 + (3+5)x + (3)(5) = x^2 + 8x + 15 \)
In simple words: Add the two numbers to get the middle term, and multiply them to get the last term.

Exam Tip: Always use the identity (x+a)(x+b) = x² + (a+b)x + ab for quick multiplication of binomials with a common x term.

 

Question 4(ii). By using standard formulae, expand the following:
\( (x+3)(x-5) \)
Answer: We know that \( (x+a)(x-b) = x^2 + (a-b)x - ab \).

\( \therefore (x+3)(x-5) = x^2 + (3-5)x - (3)(5) = x^2 - 2x - 15 \)
In simple words: Subtract the second number from the first to get the middle term, and multiply them with a minus sign for the last term.

Exam Tip: Pay careful attention to the signs - when one term is positive and one is negative, the constant term becomes negative.

 

Question 4(iii). By using standard formulae, expand the following:
\( (x-7)(x+9) \)
Answer: We know that \( (x-a)(x+b) = x^2 - (a-b)x - ab \).

\( \therefore (x-7)(x+9) = x^2 - (7-9)x - (7)(9) = x^2 + 2x - 63 \)
In simple words: Subtract the smaller value from the larger to find the middle term coefficient.

Exam Tip: When the first term is negative and the second is positive, subtract carefully to determine whether the middle term is positive or negative.

 

Question 4(iv). By using standard formulae, expand the following:
\( (x-2y)(x-3y) \)
Answer: We know that \( (x-a)(x-b) = x^2 - (a+b)x + ab \).

\( \therefore (x-2y)(x-3y) = x^2 - (2y+3y)x + (2y)(3y) = x^2 - 5xy + 6y^2 \)
In simple words: Add the two subtracted values for the middle term, and multiply them for the last term which stays positive.

Exam Tip: When both terms are negative, the product of the constants is positive, giving a positive last term in the expansion.

 

Question 5(i). By using standard formulae, expand the following:
\( (x - 2y - z)^2 \)
Answer: \( (x - 2y - z)^2 = [(x) + (-2y) + (-z)]^2 = (x)^2 + (-2y)^2 + (-z)^2 + 2[(x)(-2y) + (-2y)(-z) + (-z)(x)] \)

\( = x^2 + 4y^2 + z^2 + 2[-2xy + 2yz - xz] = x^2 + 4y^2 + z^2 - 4xy + 4yz - 2xz \)
In simple words: Square each of the three terms, then add twice the sum of all pairwise products.

Exam Tip: For trinomial squares, expand using (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca, being careful with negative signs.

 

Question 5(ii). By using standard formulae, expand the following:
\( (2x - 3y + 4z)^2 \)
Answer: \( (2x - 3y + 4z)^2 = [(2x) + (-3y) + (4z)]^2 = (2x)^2 + (-3y)^2 + (4z)^2 + 2[(2x)(-3y) + (-3y)(4z) + (4z)(2x)] \)

\( = 4x^2 + 9y^2 + 16z^2 + 2[-6xy - 12yz + 8xz] = 4x^2 + 9y^2 + 16z^2 - 12xy - 24yz + 16xz \)
In simple words: Expand by squaring each term and adding twice each pair of products, keeping track of positive and negative signs.

Exam Tip: Organize your work by listing each squared term first, then all cross-products with their correct signs to avoid mistakes.

 

Question 6(i). By using standard formulae, expand the following:
\( \left( 2x + \frac{3}{x} - 1 \right)^2 \)
Answer: \( \left( 2x + \frac{3}{x} - 1 \right)^2 = \left[ (2x) + \left(\frac{3}{x}\right) + (-1) \right]^2 = (2x)^2 + \left(\frac{3}{x}\right)^2 + (-1)^2 + 2\left[(2x)\left(\frac{3}{x}\right) + \left(\frac{3}{x}\right)(-1) + (-1)(2x)\right] \)

\( = 4x^2 + \frac{9}{x^2} + 1 + 2\left[6 - \frac{3}{x} - 2x\right] = 4x^2 + \frac{9}{x^2} + 1 + 12 - \frac{6}{x} - 4x = 4x^2 + \frac{9}{x^2} + 13 - \frac{6}{x} - 4x \)
In simple words: Treat this as a trinomial and square all three terms, then add twice the pairwise products.

Exam Tip: When fractions are involved, compute each cross-product separately and combine like terms at the end to keep the work organized.

 

Question 6(ii). By using standard formulae, expand the following:
\( \left( \frac{2}{3}x - \frac{3}{2x} - 1 \right)^2 \)
Answer: \( \left( \frac{2}{3}x - \frac{3}{2x} - 1 \right)^2 = \left[ \frac{2}{3}x + \left(-\frac{3}{2x}\right) + (-1) \right]^2 = \left(\frac{2}{3}x\right)^2 + \left(-\frac{3}{2x}\right)^2 + (-1)^2 + 2\left[\left(\frac{2}{3}x\right)\left(-\frac{3}{2x}\right) + \left(-\frac{3}{2x}\right)(-1) + (-1)\left(\frac{2}{3}x\right)\right] \)

\( = \frac{4}{9}x^2 + \frac{9}{4x^2} + 1 + 2\left[-1 + \frac{3}{2x} - \frac{2}{3}x\right] = \frac{4}{9}x^2 + \frac{9}{4x^2} + 1 - 2 + \frac{3}{x} - \frac{4}{3}x = \frac{4}{9}x^2 + \frac{9}{4x^2} - 1 + \frac{3}{x} - \frac{4}{3}x \)
In simple words: Apply the trinomial square formula carefully, computing each product and keeping track of all signs throughout.

Exam Tip: Break the calculation into stages - first square each term, then compute each cross-product separately, and combine everything at the end.

 

Question 7(i). By using standard formulae, expand the following:
\( (x+2)^3 \)
Answer: \( (x+2)^3 = x^3 + (2)^3 + 3(x)(2)(x+2) = x^3 + 8 + 6x^2 + 12x \)
In simple words: Cube the first and second terms, then add three times the product of both terms multiplied by their sum.

Exam Tip: Use the formula (a+b)³ = a³ + b³ + 3ab(a+b), which is easier than expanding step by step.

 

Question 7(ii). By using standard formulae, expand the following:
\( (2a+b)^3 \)
Answer: \( (2a+b)^3 = (2a)^3 + (b)^3 + 3(2a)(b)(2a+b) = 8a^3 + b^3 + 12a^2b + 6ab^2 \)
In simple words: Cube both parts and add three times the product of all the terms together.

Exam Tip: When the first term has a coefficient like 2a, remember that (2a)³ = 8a³, not 2a³.

 

Question 8(i). By using standard formulae, expand the following:
\( \left( 3x + \frac{1}{x} \right)^3 \)
Answer: \( \left( 3x + \frac{1}{x} \right)^3 = (3x)^3 + \left(\frac{1}{x}\right)^3 + 3(3x)\left(\frac{1}{x}\right)\left(3x + \frac{1}{x}\right) = 27x^3 + \frac{1}{x^3} + 27x + \frac{9}{x} \)
In simple words: Cube each term, add the cross-product term three times, which simplifies nicely in this case.

Exam Tip: Watch for simplifications in the cross-product term - when 3x and 1/x multiply, you get 3, which simplifies the overall expression.

 

Question 8(ii). By using standard formulae, expand the following:
\( (2x-1)^3 \)
Answer: \( (2x-1)^3 = (2x)^3 - (1)^3 - 3(2x)(1)(2x-1) = 8x^3 - 1 - 12x^2 + 6x \)
In simple words: Cube both terms and subtract three times the product of both terms multiplied by their difference.

Exam Tip: For (a-b)³, use a³ - b³ - 3ab(a-b), ensuring that you handle the negative signs correctly throughout.

 

Question 9(i). By using standard formulae, expand the following:
\( (5x-3y)^3 \)
Answer: \( (5x-3y)^3 = (5x)^3 - (3y)^3 - 3(5x)(3y)(5x-3y) = 125x^3 - 27y^3 - 225x^2y + 135xy^2 \)
In simple words: Cube the first term, subtract the cube of the second term, then subtract the cross-product term.

Exam Tip: Always expand the cross-product term carefully - here 3(5x)(3y)(5x-3y) gives two separate terms after distribution.

 

Question 9(ii). By using standard formulae, expand the following:
\( \left( 2x - \frac{1}{3y} \right)^3 \)
Answer: \( \left( 2x - \frac{1}{3y} \right)^3 = (2x)^3 - \left(\frac{1}{3y}\right)^3 - 3(2x)\left(\frac{1}{3y}\right)\left(2x - \frac{1}{3y}\right) = 8x^3 - \frac{1}{27y^3} - \frac{4x^2}{y} + \frac{2x}{3y^2} \)
In simple words: Apply the difference-of-cubes expansion, being careful with fraction exponents and products.

Exam Tip: When working with cubic fractions, compute each term separately before combining to avoid arithmetic errors.

 

Question 10(i). Simplify the following:
\( \left( 2x - \frac{1}{3y} \right)^3 \)
Answer: \( \left( 2x - \frac{1}{3y} \right)^3 = (2x)^3 - \left(\frac{1}{3y}\right)^3 - 3(2x)\left(\frac{1}{3y}\right)\left(2x - \frac{1}{3y}\right) = 8x^3 - \frac{1}{27y^3} - \frac{4x^2}{y} + \frac{2x}{3y^2} \)
In simple words: Use the cubic expansion formula for negative binomials to get the final simplified form.

Exam Tip: Organize work by computing each cubic and cross-product term before combining all parts into the final answer.

 

Question 10(ii). Simplify the following:
\( \left( a + \frac{1}{a} \right)^2 + \left( a - \frac{1}{a} \right)^2 \)
Answer: \( \left( a + \frac{1}{a} \right)^2 + \left( a - \frac{1}{a} \right)^2 = \left[ a^2 + 2 + \frac{1}{a^2} \right] + \left[ a^2 - 2 + \frac{1}{a^2} \right] = 2a^2 + \frac{2}{a^2} = 2\left( a^2 + \frac{1}{a^2} \right) \)
In simple words: Expand each square separately using the binomial formula, then add the results to see the middle terms cancel.

Exam Tip: Notice that when adding (a+1/a)² and (a-1/a)², the cross-terms cancel out because one is +2 and the other is -2.

 

Question 10(ii) (continued). Simplify the following:
\( \left( a + \frac{1}{a} \right)^2 - \left( a - \frac{1}{a} \right)^2 \)
Answer: \( \left( a + \frac{1}{a} \right)^2 - \left( a - \frac{1}{a} \right)^2 = \left[ a^2 + 2 + \frac{1}{a^2} \right] - \left[ a^2 - 2 + \frac{1}{a^2} \right] = a^2 + 2 + \frac{1}{a^2} - a^2 + 2 - \frac{1}{a^2} = 4 \)
In simple words: Expand both expressions and subtract carefully - the squared terms cancel, leaving only the constant 4.

Exam Tip: When subtracting one expansion from another, distribute the negative sign to all terms of the second expression before simplifying.

 

Question 11(i). Simplify the following:
\( (3x-1)^2 - (3x-2)(3x+1) \)
Answer: \( (3x-1)^2 - (3x-2)(3x+1) = [(3x)^2 - 2(3x)(1) + (1)^2] - [(3x)^2 + 3x - 6x - 2] = (9x^2 - 6x + 1) - (9x^2 - 3x - 2) = 9x^2 - 6x + 1 - 9x^2 + 3x + 2 = -3x + 3 = 3 - 3x \)
In simple words: Expand each part separately, then subtract the second expanded form from the first, combining like terms.

Exam Tip: When subtracting expressions, distribute the negative sign to every term of the second bracket before combining like terms.

 

Question 11(ii). Simplify the following:
\( (4x+3y)^2 - (4x-3y)^2 - 48xy \)
Answer: \( (4x+3y)^2 - (4x-3y)^2 - 48xy = [16x^2 + 24xy + 9y^2] - [16x^2 - 24xy + 9y^2] - 48xy = 16x^2 + 24xy + 9y^2 - 16x^2 + 24xy - 9y^2 - 48xy = 48xy - 48xy = 0 \)
In simple words: The squared terms cancel when subtracting, leaving only cross-product terms that also cancel when you subtract the third term.

Exam Tip: Observe that this expression equals zero - always look for such patterns where terms cancel completely.

 

Question 12(i). Simplify the following:
\( (7p+9q)(7p-9q) \)
Answer: \( (7p+9q)(7p-9q) = (7p)^2 - (9q)^2 = 49p^2 - 81q^2 \)
In simple words: Use the difference-of-squares formula: the product of a sum and difference equals the first term squared minus the second term squared.

Exam Tip: Whenever you see (a+b)(a-b), immediately apply the identity (a+b)(a-b) = a² - b² for a quick answer.

 

Question 12(ii). Simplify the following:
\( \left( 2x - \frac{3}{x} \right)\left( 2x + \frac{3}{x} \right) \)
Answer: \( \left( 2x - \frac{3}{x} \right)\left( 2x + \frac{3}{x} \right) = (2x)^2 - \left(\frac{3}{x}\right)^2 = 4x^2 - \frac{9}{x^2} \)
In simple words: This is a sum-times-difference pattern, so square the first term and subtract the square of the second term.

Exam Tip: Always look for the (a-b)(a+b) = a² - b² pattern - it simplifies multiplication instantly.

 

Question 13(i). Simplify the following:
\( (2x - y + 3)(2x - y - 3) \)
Answer: Let \( (2x - y) = a \). Then the given expression = \( (a+3)(a-3) = a^2 - 9 = (2x-y)^2 - 9 = (2x)^2 - 2(2x)(y) + y^2 - 9 = 4x^2 - 4xy + y^2 - 9 \)
In simple words: Recognize that 2x-y appears in both parts, so set it equal to a, apply the difference-of-squares formula, then expand.

Exam Tip: When a common expression appears in two factors, use substitution to simplify before expanding further.

 

Question 13(ii). Simplify the following:
\( (3x+y-5)(3x-y-5) \)
Answer: \( (3x+y-5)(3x-y-5) = (3x-5+y)(3x-5-y) \). Let \( (3x-5) = a \). Then the given expression = \( (a+y)(a-y) = a^2 - y^2 = (3x-5)^2 - y^2 = [(3x)^2 - 2(3x)(5) + (5)^2] - y^2 = 9x^2 - 30x + 25 - y^2 \)
In simple words: Rearrange to find a common part, substitute a variable, apply the difference-of-squares formula, then expand.

Exam Tip: Rearranging terms can reveal hidden patterns - here grouping 3x and -5 together makes the difference-of-squares pattern clear.

 

Question 14(i). Simplify the following:
\( \left( x + \frac{2}{x} - 3 \right)\left( x - \frac{2}{x} - 3 \right) \)
Answer: Let \( (x - 3) = a \). Then the given expression = \( \left( a + \frac{2}{x} \right)\left( a - \frac{2}{x} \right) = a^2 - \left(\frac{2}{x}\right)^2 = (x-3)^2 - \frac{4}{x^2} = x^2 - 2(x)(3) + 3^2 - \frac{4}{x^2} = x^2 - 6x + 9 - \frac{4}{x^2} \)
In simple words: Group x-3 as a single unit, then use the difference-of-squares formula with the remaining 2/x terms.

Exam Tip: Strategic substitution helps reveal the underlying algebraic pattern and makes complex-looking expressions much simpler to handle.

 

Question 14(ii). Simplify the following:
\( (5-2x)(5+2x)(25+4x^2) \)
Answer: \( (5-2x)(5+2x)(25+4x^2) = [(5)^2 - (2x)^2](25+4x^2) = (25-4x^2)(25+4x^2) = (25)^2 - (4x^2)^2 = 625 - 16x^4 \)
In simple words: First use difference-of-squares on the first two factors, then apply it again to the result with the third factor.

Exam Tip: Look for patterns where you can apply the same formula twice - here the second application uses a² - b² again with a=25 and b=4x².

 

Question 15(i). Simplify the following:
\( (x+2y+3)(x+2y+7) \)
Answer: Let \( (x+2y) = a \). Then the given expression = \( (a+3)(a+7) = a^2 + (3+7)a + (3)(7) = a^2 + 10a + 21 = (x+2y)^2 + 10(x+2y) + 21 = x^2 + 4xy + 4y^2 + 10x + 20y + 21 \)
In simple words: Substitute a common binomial with a variable, use the binomial expansion formula, then substitute back and expand completely.

Exam Tip: When the same expression appears in both factors with different constants, use substitution and the (x+a)(x+b) = x² + (a+b)x + ab formula.

 

Question 15(ii). Simplify the following:
\( (2x+y+5)(2x+y-9) \)
Answer: Let \( (2x+y)=z \). Then the expression becomes \( (z+5)(z-9) \).

Using the identity \( (x+a)(x-b) = x^2 + (a-b)x - ab \):

\( (z+5)(z-9) = z^2 + (5-9)z - (5)(9) \)

\( = z^2 - 4z - 45 \)

\( = (2x+y)^2 - 4(2x+y) - 45 \)

\( = 4x^2 + 4xy + y^2 - 8x - 4y - 45 \)

\( = 4x^2 - 8x + y^2 - 4y + 4xy - 45 \)
In simple words: Replace the common part with a single letter to make the expression shorter. Then apply the multiplication rule and expand it back.

Exam Tip: Always identify and substitute the repeated expression first, then expand the result completely by replacing the substituted variable back into the original form.

 

Question 15(iii). Simplify the following:
\( (x - 2y - 5)(x - 2y + 3) \)
Answer: Let \( (x-2y) = z \). Then the expression becomes \( (z-5)(z+3) \).

Using the identity \( (x-a)(x+b) = x^2 - (a-b)x - ab \):

\( (z-5)(z+3) = z^2 - (5-3)z - (5)(3) \)

\( = z^2 - 2z - 15 \)

\( = (x-2y)^2 - 2(x-2y) - 15 \)

\( = x^2 - 4xy + 4y^2 - 2x + 4y - 15 \)
In simple words: Find the matching part in both brackets, replace it with one letter, multiply using the rule, then substitute back and simplify.

Exam Tip: Pay attention to the signs inside the brackets when choosing your substitution variable, as this affects the final formula you apply.

 

Question 15(iv). Simplify the following:
\( (3x - 4y - 2)(3x - 4y - 6) \)
Answer: Let \( (3x-4y) = z \). Then the expression becomes \( (z-2)(z-6) \).

Using the identity \( (x-a)(x-b) = x^2 - (a+b)x + ab \):

\( (z-2)(z-6) = z^2 - (2+6)z + (2)(6) \)

\( = z^2 - 8z + 12 \)

\( = (3x-4y)^2 - 8(3x-4y) + 12 \)

\( = 9x^2 - 24xy + 16y^2 - 24x + 32y + 12 \)
In simple words: Take out the shared part as a single variable. Use the formula for two identical starting parts with different endings. Expand and rewrite using the original variables.

Exam Tip: Double-check your expansion of the squared term - use \( (a-b)^2 = a^2 - 2ab + b^2 \) carefully to avoid sign errors.

 

Question 16(i). Simplify the following:
\( (2p + 3q)(4p^2 - 6pq + 9q^2) \)
Answer: Notice that \( (2p + 3q)(4p^2 - 6pq + 9q^2) = (2p + 3q)[(2p)^2 - (2p)(3q) + (3q)^2] \).

This matches the identity \( (a + b)(a^2 - ab + b^2) = a^3 + b^3 \), where \( a = 2p \) and \( b = 3q \).

Applying this:
\( (2p + 3q)[(2p)^2 - (2p)(3q) + (3q)^2] = (2p)^3 + (3q)^3 \)

\( = 8p^3 + 27q^3 \)
In simple words: When you see a binomial times a trinomial with this special pattern, it gives you the sum of cubes. Just cube each part of the binomial separately.

Exam Tip: Recognise the pattern early - if the trinomial has the form \( a^2 - ab + b^2 \) paired with \( (a+b) \), you get \( a^3 + b^3 \) immediately without expanding.

 

Question 16(ii). Simplify the following:
\( \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right) \)
Answer: Rewrite the second bracket: \( x^2 - 1 + \frac{1}{x^2} = \left(x\right)^2 - \left(x\right)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 \).

Using the identity \( (a + b)(a^2 - ab + b^2) = a^3 + b^3 \) with \( a = x \) and \( b = \frac{1}{x} \):

\( \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right) = x^3 + \frac{1}{x^3} \)
In simple words: Check if the trinomial fits the sum of cubes pattern with the binomial. If it does, the answer is simply the cube of each term from the binomial.

Exam Tip: Always simplify the middle term of the trinomial first to see if it matches the \( -ab \) pattern you need.

 

Question 17(i). Simplify the following:
\( (3p - 4q)(9p^2 + 12pq + 16q^2) \)
Answer: Notice that \( (3p - 4q)(9p^2 + 12pq + 16q^2) = (3p - 4q)[(3p)^2 + (3p)(4q) + (4q)^2] \).

This matches the identity \( (a - b)(a^2 + ab + b^2) = a^3 - b^3 \), where \( a = 3p \) and \( b = 4q \).

Applying this:
\( (3p - 4q)[(3p)^2 + (3p)(4q) + (4q)^2] = (3p)^3 - (4q)^3 \)

\( = 27p^3 - 64q^3 \)
In simple words: When a binomial with a minus sign is multiplied by a trinomial with this special pattern, you get the difference of cubes. Cube each part separately and subtract.

Exam Tip: Spot the difference of cubes pattern early: \( (a-b) \) with \( (a^2 + ab + b^2) \) gives \( a^3 - b^3 \) directly.

 

Question 17(ii). Simplify the following:
\( \left(x - \frac{3}{x}\right)\left(x^2 + 3 + \frac{9}{x^2}\right) \)
Answer: Rewrite the second bracket: \( x^2 + 3 + \frac{9}{x^2} = \left(x\right)^2 + \left(x\right)\left(\frac{3}{x}\right) + \left(\frac{3}{x}\right)^2 \).

Using the identity \( (a - b)(a^2 + ab + b^2) = a^3 - b^3 \) with \( a = x \) and \( b = \frac{3}{x} \):

\( \left(x - \frac{3}{x}\right)\left(x^2 + 3 + \frac{9}{x^2}\right) = x^3 - \left(\frac{3}{x}\right)^3 \)

\( = x^3 - \frac{27}{x^3} \)
In simple words: If the trinomial has the pattern \( a^2 + ab + b^2 \) paired with \( (a-b) \), then cube each term from the binomial and subtract to get your answer.

Exam Tip: Verify that the middle term of the trinomial equals the product of the two terms in the binomial - this confirms the identity applies.

 

Question 18. Simplify the following:
\( (2x + 3y + 4z)(4x^2 + 9y^2 + 16z^2 - 6xy - 12yz - 8zx) \)
Answer: Notice that \( (2x + 3y + 4z) \) can be rewritten, and the second part follows the pattern:
\( (4x^2 + 9y^2 + 16z^2 - 6xy - 12yz - 8zx) = (2x)^2 + (3y)^2 + (4z)^2 - (2x)(3y) - (3y)(4z) - (4z)(2x) \).

Using the identity \( (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \) with \( a = 2x \), \( b = 3y \), \( c = 4z \):

\( = (2x)^3 + (3y)^3 + (4z)^3 - 3(2x)(3y)(4z) \)

\( = 8x^3 + 27y^3 + 64z^3 - 72xyz \)
In simple words: This formula works for three terms. Cube each one, add them together, then subtract three times the product of all three.

Exam Tip: Always check that the trinomial matches the full pattern including all six terms - missing or incorrect signs mean the identity does not apply.

 

Question 19(i). Find the product of the following:
\( (x + 1)(x + 2)(x + 3) \)
Answer: Use the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ca)x + abc \).

With \( a = 1 \), \( b = 2 \), \( c = 3 \):

\( (x + 1)(x + 2)(x + 3) = x^3 + (1 + 2 + 3)x^2 + [(1)(2) + (2)(3) + (3)(1)]x + (1)(2)(3) \)

\( = x^3 + 6x^2 + (2 + 6 + 3)x + 6 \)

\( = x^3 + 6x^2 + 11x + 6 \)
In simple words: Add all three numbers to get the \( x^2 \) coefficient. Add all three pair products to get the \( x \) coefficient. Multiply all three to get the constant term.

Exam Tip: Carefully compute the pair products: \( ab \), \( bc \), and \( ca \) - missing any one leads to an incorrect middle term.

 

Question 19(ii). Find the product of the following:
\( (x - 2)(x - 3)(x + 4) \)
Answer: Use the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ca)x + abc \).

With \( a = -2 \), \( b = -3 \), \( c = 4 \):

\( (x - 2)(x - 3)(x + 4) = x^3 + [(-2) + (-3) + 4]x^2 + [(-2)(-3) + (-3)(4) + (4)(-2)]x + (-2)(-3)(4) \)

\( = x^3 - x^2 + (6 - 12 - 8)x + 24 \)

\( = x^3 - x^2 - 14x + 24 \)
In simple words: Treat each number with its sign. Add them for the \( x^2 \) term. Multiply pairs (watching signs) and add for the \( x \) term. Multiply all three for the number at the end.

Exam Tip: Watch your signs carefully when \( a \), \( b \), and \( c \) are negative - a common place to make mistakes is in computing \( ab + bc + ca \).

 

Question 20. Find the coefficient of x² and x in the product of (x - 3)(x + 7)(x - 4)
Answer: Using the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ca)x + abc \).

Comparing with \( (x - 3)(x + 7)(x - 4) \), we have \( a = -3 \), \( b = 7 \), \( c = -4 \).

Coefficient of \( x^2 \) is \( a + b + c = (-3) + 7 + (-4) = 0 \).

Coefficient of \( x \) is \( ab + bc + ca = (-3)(7) + (7)(-4) + (-4)(-3) = -21 - 28 + 12 = -37 \).
In simple words: The coefficient of \( x^2 \) comes from adding the three numbers. The coefficient of \( x \) comes from adding the products of every pair of numbers.

Exam Tip: It is easy to miss a pair product or make an arithmetic error - write out all three pairs and calculate each one separately before adding.

 

Question 21. If a² + 4a + x = (a + 2)², find the value of x
Answer: Given: \( a^2 + 4a + x = (a + 2)^2 \).

Expanding the right side using \( (a + 2)^2 = a^2 + 2(a)(2) + 2^2 \):

\( a^2 + 4a + x = a^2 + 4a + 4 \)

Subtracting \( a^2 + 4a \) from both sides:

\( x = 4 \)
In simple words: Expand the right side completely, then compare both sides to see what value is missing on the left side.

Exam Tip: Always expand the perfect square on the right side term-by-term - this makes it easy to match and find the missing value.

 

Question 22(i). Use (a + b)² = a² + 2ab + b² to evaluate the following: (101)²
Answer: Write \( 101 = 100 + 1 \).

Using \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( a = 100 \) and \( b = 1 \):

\( (100 + 1)^2 = (100)^2 + 2(100)(1) + (1)^2 \)

\( = 10000 + 200 + 1 = 10201 \)
In simple words: Break the number into a round part and a small part. Square each, double their product, and add everything together.

Exam Tip: Choose your split wisely - use numbers close to powers of 10 or other easy-to-square values to make the arithmetic fast.

 

Question 22(ii). Use (a + b)² = a² + 2ab + b² to evaluate the following: (1003)²
Answer: Write \( 1003 = 1000 + 3 \).

Using \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( a = 1000 \) and \( b = 3 \):

\( (1000 + 3)^2 = (1000)^2 + 2(1000)(3) + (3)^2 \)

\( = 1000000 + 6000 + 9 = 1006009 \)
In simple words: Separate the number into a larger round part and a small part. Apply the formula and add the three results.

Exam Tip: Multiplying by 1000 or higher powers makes the arithmetic cleaner - split accordingly to avoid large intermediate calculations.

 

Question 22(iii). Use (a + b)² = a² + 2ab + b² to evaluate the following: (10.2)²
Answer: Write \( 10.2 = 10 + 0.2 \).

Using \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( a = 10 \) and \( b = 0.2 \):

\( (10 + 0.2)^2 = (10)^2 + 2(10)(0.2) + (0.2)^2 \)

\( = 100 + 4 + 0.04 = 104.04 \)
In simple words: Split the decimal into a whole number and a fractional part. Use the formula with these parts and add the three terms.

Exam Tip: Be careful with decimal arithmetic - \( 2(10)(0.2) = 4 \) and \( (0.2)^2 = 0.04 \), not 0.4.

 

Question 23(i). Use (a - b)² = a² - 2ab + b² to evaluate the following: (99)²
Answer: Write \( 99 = 100 - 1 \).

Using \( (a - b)^2 = a^2 - 2ab + b^2 \) with \( a = 100 \) and \( b = 1 \):

\( (100 - 1)^2 = (100)^2 - 2(100)(1) + (1)^2 \)

\( = 10000 - 200 + 1 = 9801 \)
In simple words: Express the number as a larger round value minus a smaller one. Apply the formula - note that the middle term is subtracted, not added.

Exam Tip: Remember the minus sign in the middle term - this is the key difference from the \( (a+b)^2 \) formula and a common source of errors.

 

Question 23(ii). Use (a - b)² = a² - 2ab + b² to evaluate the following: (997)²
Answer: Write \( 997 = 1000 - 3 \).

Using \( (a - b)^2 = a^2 - 2ab + b^2 \) with \( a = 1000 \) and \( b = 3 \):

\( (1000 - 3)^2 = (1000)^2 - 2(1000)(3) + (3)^2 \)

\( = 1000000 - 6000 + 9 = 994009 \)
In simple words: Break the number into a round thousand minus a small value. Use the formula, being careful with the subtraction in the middle.

Exam Tip: The final answer is close to the larger base squared - use this as a quick check that your subtraction and addition have been done correctly.

 

Question 23(iii). Use (a - b)² = a² - 2ab + b² to evaluate the following: (9.8)²
Answer: Write \( 9.8 = 10 - 0.2 \).

Using \( (a - b)^2 = a^2 - 2ab + b^2 \) with \( a = 10 \) and \( b = 0.2 \):

\( (10 - 0.2)^2 = (10)^2 - 2(10)(0.2) + (0.2)^2 \)

\( = 100 - 4 + 0.04 = 96.04 \)
In simple words: Split the decimal by expressing it as a round number minus a small decimal part. Apply the difference of squares formula, subtracting the middle term.

Exam Tip: Handle decimals carefully - \( 2(10)(0.2) = 4 \) exactly, which cancels nicely, leaving \( 100 - 4 + 0.04 \).

 

Question 24(i). By using suitable identities, evaluate the following: (103)³
Answer: Write \( 103 = 100 + 3 \).

Using \( (a + b)^3 = a^3 + b^3 + 3(a)(b)(a + b) \) with \( a = 100 \) and \( b = 3 \):

\( (100 + 3)^3 = (100)^3 + (3)^3 + 3(100)(3)(100 + 3) \)

\( = 1000000 + 27 + 900(103) \)

\( = 1000000 + 27 + 92700 = 1092727 \)
In simple words: Cube the first part, cube the second part, then add three times their product times their sum. Add all three results.

Exam Tip: Calculate the coefficient first - here \( 3(100)(3) = 900 \) - then multiply it by the sum to get the third term.

 

Question 24(ii). By using suitable identities, evaluate the following: (99)³
Answer: Write \( 99 = 100 - 1 \).

Using \( (a - b)^3 = a^3 - b^3 - 3(a)(b)(a - b) \) with \( a = 100 \) and \( b = 1 \):

\( (100 - 1)^3 = (100)^3 - (1)^3 - 3(100)(1)(100 - 1) \)

\( = 1000000 - 1 - 300(99) \)

\( = 1000000 - 1 - 29700 = 970299 \)
In simple words: Cube the first part, cube the second part, then subtract three times their product times their difference. Subtract both from the first cube.

Exam Tip: Watch the signs - both the second cube and the third term are subtracted from the first cube in this formula.

 

Question 24(iii). By using suitable identities, evaluate the following: (10.1)³
Answer: Write \( 10.1 = 10 + 0.1 \).

Using \( (a + b)^3 = a^3 + b^3 + 3(a)(b)(a + b) \) with \( a = 10 \) and \( b = 0.1 \):

\( (10 + 0.1)^3 = (10)^3 + (0.1)^3 + 3(10)(0.1)(10 + 0.1) \)

\( = 1000 + 0.001 + 3(10.1) \)

\( = 1000 + 0.001 + 30.3 = 1030.301 \)
In simple words: Split the decimal into integer and fractional parts. Cube each, add their product term, and combine the three parts together.

Exam Tip: Be precise with small decimals - \( (0.1)^3 = 0.001 \) and \( 3(10)(0.1) = 3 \), which multiplied by 10.1 gives 30.3.

 

Question 25. If 2a - b + c = 0, prove 4a² - b² + c² + 4ac = 0
Answer: Given: \( 2a - b + c = 0 \).

Rearrange: \( 2a + c = b \).

Square both sides: \( (2a + c)^2 = b^2 \).

Expand: \( 4a^2 + 4ac + c^2 = b^2 \).

Rearrange: \( 4a^2 + c^2 + 4ac - b^2 = 0 \).

This is the same as: \( 4a^2 - b^2 + c^2 + 4ac = 0 \). Hence proved.
In simple words: Move the \( b \) term to one side of the equation. Square both sides to get a squared equation. Rearrange to match what you need to prove.

Exam Tip: Always show the expansion step-by-step - examiners look for evidence that you understand the perfect square formula, not just that you guessed the answer.

 

Question 26. If a + b + 2c = 0, prove that a³ + b³ + 8c³ = 6abc
Answer: Given: \( a + b + 2c = 0 \).

We know that if \( p + q + r = 0 \), then \( p^3 + q^3 + r^3 = 3pqr \).

Applying this identity with \( p = a \), \( q = b \), \( r = 2c \):

\( a^3 + b^3 + (2c)^3 = 3(a)(b)(2c) \)

\( a^3 + b^3 + 8c^3 = 6abc \)

Hence proved.
In simple words: Notice that the given equation has the form "sum equals zero" with three parts. Use the identity for that form, treating \( 2c \) as a single quantity.

Exam Tip: When you see a cubic identity to prove and a condition like "sum equals zero," think immediately of the \( a^3 + b^3 + c^3 - 3abc \) identity.

 

Question 27. If x + 2y - 3 = 0, then find the value of x³ + 8y³ + 6x²y + 12xy² - 125.
Answer: Given: \( x + 2y - 3 = 0 \), so \( x + 2y = 3 \).

Observe that \( x^3 + 8y^3 + 6x^2y + 12xy^2 = x^3 + (2y)^3 + 6xy(x + 2y) \).

Factor further: \( x^3 + (2y)^3 + 3 \cdot x \cdot 2y \cdot (x + 2y) = (x + 2y)^3 \).

Substitute \( x + 2y = 3 \):

\( (x + 2y)^3 - 125 = 3^3 - 125 = 27 - 125 = -98 \)

Therefore, the value is \( -98 \).
In simple words: Group the terms so they match the cube of a sum formula. Recognise that the remaining terms form \( (x+2y)^3 \). Substitute the given value and simplify.

Exam Tip: Spot the pattern in the coefficients - the 6 and 12 suggest expansion of a cubic, and knowing the form helps you group and simplify quickly.

 

Question 28. If a + b + c = 0, then find the value of \( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \)
Answer: Given: \( a + b + c = 0 \).

We know that if \( a + b + c = 0 \), then \( a^3 + b^3 + c^3 = 3abc \).

Divide both sides by \( abc \):

\( \frac{a^3}{abc} + \frac{b^3}{abc} + \frac{c^3}{abc} = \frac{3abc}{abc} \)

\( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3 \)
In simple words: Start with the identity for "sum equals zero." Divide the whole equation by the product \( abc \) to transform the cubes into the fractions you need.

Exam Tip: Always recall that \( a + b + c = 0 \) implies \( a^3 + b^3 + c^3 = 3abc \) - this is a high-frequency identity used in many proof questions.

 

Question 29. If x + y = 4, find the value of x³ + y³ + 12xy - 64
Answer: Using the identity \( (x + y)^3 = x^3 + y^3 + 3(x)(y)(x + y) \)

When we substitute x + y = 4:

\( x^3 + y^3 + 3xy(4) = (4)^3 \)

\( \implies x^3 + y^3 + 12xy = 64 \)

\( \implies x^3 + y^3 + 12xy - 64 = 0 \)

In simple words: When x + y equals 4, the expression x³ + y³ + 12xy always equals 64, so subtracting 64 gives 0.

Exam Tip: Recognize the cube expansion formula and match the form (x + y)³ immediately - this saves calculation time and avoids errors.

 

Question 30. Without actually calculating the cubes, find the values of:
(i) \( (27)^3 + (-17)^3 + (-10)^3 \)
(ii) \( (-28)^3 + (15)^3 + (13)^3 \)
Answer:
(i) Setting a = 27, b = -17, c = -10

We observe that \( a + b + c = 27 - 17 - 10 = 0 \)

Using the identity: when a + b + c = 0, then \( a^3 + b^3 + c^3 = 3abc \)

\( \implies (27)^3 + (-17)^3 + (-10)^3 = 3(27)(-17)(-10) = 13770 \)

(ii) Setting a = -28, b = 15, c = 13

We observe that \( a + b + c = -28 + 15 + 13 = 0 \)

Using the same identity:

\( \implies (-28)^3 + (15)^3 + (13)^3 = 3(-28)(15)(13) = -16380 \)

In simple words: When three numbers add up to zero, their cubes add up to 3 times their product. We don't need to cube anything - just multiply the three numbers and multiply by 3.

Exam Tip: Always check if a + b + c = 0 before starting; this identity cuts calculation time dramatically and is a favorite trick for avoiding tedious arithmetic.

 

Question 31. Using suitable identity, find the value of: \( \frac{86 \times 86 \times 86 + 14 \times 14 \times 14}{86 \times 86 - 86 \times 14 + 14 \times 14} \)
Answer: Let x = 86 and y = 14.

The given expression becomes:

\( \frac{x^3 + y^3}{x^2 - xy + y^2} \)

Using the factorization identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \):

\( \frac{(x + y)(x^2 - xy + y^2)}{x^2 - xy + y^2} = x + y \)

The terms \( (x^2 - xy + y^2) \) cancel from numerator and denominator:

\( = 86 + 14 = 100 \)

In simple words: The numerator factors as (x + y) times (x² - xy + y²), so these pieces cancel with the denominator, leaving just x + y.

Exam Tip: Spot the sum of cubes pattern immediately - it factors into a product that will cancel with the denominator, making the problem much simpler than it looks.

 

Exercise 3.2

 

Question 1. If x - y = 8 and xy = 5, find x² + y².
Answer: We know the identity \( (x - y)^2 = x^2 - 2xy + y^2 \)

Rearranging: \( x^2 + y^2 = (x - y)^2 + 2xy \)

Substituting the given values:

\( x^2 + y^2 = (8)^2 + 2(5) = 64 + 10 = 74 \)

In simple words: Square the difference, then add twice the product. That gives the sum of squares.

Exam Tip: Remember this formula - it appears constantly. Practice rearranging (x - y)² to build fluency with these identities.

 

Question 2. If x + y = 10 and xy = 21, find 2(x² + y²).
Answer: We use the identity \( (x + y)^2 = x^2 + 2xy + y^2 \)

Rearranging: \( x^2 + y^2 = (x + y)^2 - 2xy \)

Therefore: \( 2(x^2 + y^2) = 2[(x + y)^2 - 2xy] \)

Substituting the values:

\( 2(x^2 + y^2) = 2[(10)^2 - 2(21)] = 2[100 - 42] = 2(58) = 116 \)

In simple words: Square the sum, subtract twice the product, then multiply everything by 2.

Exam Tip: Double-check your arithmetic at each step - mistakes in multiplying by 2 at the end are common.

 

Question 3. If 2a + 3b = 7 and ab = 2, find 4a² + 9b².
Answer: We recognize that \( 4a^2 + 9b^2 = (2a)^2 + (3b)^2 \)

Using the identity \( x^2 + y^2 = (x + y)^2 - 2xy \) where x = 2a and y = 3b:

\( 4a^2 + 9b^2 = (2a + 3b)^2 - 2(2a)(3b) = (2a + 3b)^2 - 12ab \)

Substituting the given values:

\( 4a^2 + 9b^2 = (7)^2 - 12(2) = 49 - 24 = 25 \)

In simple words: Notice that 4a² is really (2a) squared and 9b² is really (3b) squared. Then use the standard sum-of-squares formula.

Exam Tip: Always rewrite coefficients as perfect squares first (4 = 2², 9 = 3²) so you can apply the standard formulas.

 

Question 4. If 3x - 4y = 16 and xy = 4, find the value of 9x² + 16y².
Answer: We recognize that \( 9x^2 + 16y^2 = (3x)^2 + (4y)^2 \)

Using the identity \( x^2 + y^2 = (x - y)^2 + 2xy \) where x = 3x and y = 4y:

\( 9x^2 + 16y^2 = (3x - 4y)^2 + 2(3x)(4y) = (3x - 4y)^2 + 24xy \)

Substituting the values:

\( 9x^2 + 16y^2 = (16)^2 + 24(4) = 256 + 96 = 352 \)

In simple words: Rewrite 9x² as (3x)² and 16y² as (4y)², then apply the identity for the sum of squares with a difference term.

Exam Tip: Notice the pattern - whenever you see coefficients 4, 9, 16, 25, etc., think of them as perfect squares and rewrite using factored forms.

 

Question 5. If x + y = 8 and x - y = 2, find the value of 2x² + 2y².
Answer: We use two basic identities:

\( (x + y)^2 = x^2 + y^2 + 2xy \) ... (i)

\( (x - y)^2 = x^2 + y^2 - 2xy \) ... (ii)

Adding equations (i) and (ii):

\( (x + y)^2 + (x - y)^2 = 2x^2 + 2y^2 \)

Therefore: \( 2x^2 + 2y^2 = (x + y)^2 + (x - y)^2 \)

Substituting the values:

\( 2x^2 + 2y^2 = (8)^2 + (2)^2 = 64 + 4 = 68 \)

In simple words: Add the squares of the sum and the difference. This gives twice the sum of squares.

Exam Tip: When both (x + y) and (x - y) are given, immediately add their squares - this is faster than finding x and y individually.

 

Question 6. If a² + b² = 13 and ab = 6, find
(i) \( a + b \)
(ii) \( a - b \)
Answer:
(i) Using \( (a + b)^2 = a^2 + b^2 + 2ab \):

\( (a + b)^2 = 13 + 2(6) = 13 + 12 = 25 \)

\( a + b = \pm 5 \)

(ii) Using \( (a - b)^2 = a^2 + b^2 - 2ab \):

\( (a - b)^2 = 13 - 2(6) = 13 - 12 = 1 \)

\( a - b = \pm 1 \)

In simple words: Both the sum and difference formulas let you find (a + b) and (a - b) without finding a and b separately. Take square roots to finish.

Exam Tip: Remember both ± values are valid unless the problem specifies positive or negative variables - show both roots.

 

Question 7. If a + b = 4 and ab = -12, find
(i) \( a - b \)
(ii) \( a^2 - b^2 \)
Answer:
(i) Using the identity \( (a - b)^2 = (a + b)^2 - 4ab \):

\( (a - b)^2 = (4)^2 - 4(-12) = 16 + 48 = 64 \)

\( a - b = \pm 8 \)

(ii) Using the factorization \( a^2 - b^2 = (a + b)(a - b) \):

\( a^2 - b^2 = 4 \times (\pm 8) = \pm 32 \)

In simple words: First find a - b using the rearranged (a + b)² formula. Then multiply (a + b) by (a - b) to get the difference of squares.

Exam Tip: The ± sign propagates through to part (ii) - show both cases clearly to avoid losing marks.

 

Question 8. If p - q = 9 and pq = 36, evaluate
(i) \( p + q \)
(ii) \( p^2 - q^2 \)
Answer:
(i) Using the rearranged identity \( (p + q)^2 = (p - q)^2 + 4pq \):

\( (p + q)^2 = (9)^2 + 4(36) = 81 + 144 = 225 \)

\( p + q = \pm 15 \)

(ii) Using \( p^2 - q^2 = (p - q)(p + q) \):

\( p^2 - q^2 = 9 \times (\pm 15) = \pm 135 \)

In simple words: When p - q is given, rearrange the (p + q)² formula to use (p - q)² instead. Then use the difference of squares factorization.

Exam Tip: This type reverses the usual pattern - the difference is given instead of the sum. Stay flexible with identity rearrangement.

 

Question 9. If x + y = 6 and x - y = 4, find
(i) \( x^2 + y^2 \)
(ii) \( xy \)
Answer:
(i) Using the identities \( (x + y)^2 = x^2 + y^2 + 2xy \) and \( (x - y)^2 = x^2 + y^2 - 2xy \):

Adding them:

\( (x + y)^2 + (x - y)^2 = 2x^2 + 2y^2 \)

Therefore: \( x^2 + y^2 = \frac{(x + y)^2 + (x - y)^2}{2} \)

\( x^2 + y^2 = \frac{(6)^2 + (4)^2}{2} = \frac{36 + 16}{2} = \frac{52}{2} = 26 \)

(ii) Subtracting the two equations:

\( (x + y)^2 - (x - y)^2 = 4xy \)

Therefore: \( xy = \frac{(x + y)^2 - (x - y)^2}{4} \)

\( xy = \frac{(6)^2 - (4)^2}{4} = \frac{36 - 16}{4} = \frac{20}{4} = 5 \)

In simple words: When both the sum and difference are given, add their squares to find x² + y², and subtract them to find xy. Split the result by 2 or 4 as needed.

Exam Tip: These are efficient shortcuts - much faster than solving for x and y individually when only their sum and difference are known.

 

Question 10. If \( x - 3 = \frac{1}{x} \), find the value of \( x^2 + \frac{1}{x^2} \).
Answer: From the given equation \( x - 3 = \frac{1}{x} \), we get:

\( x - \frac{1}{x} = 3 \)

Using the identity \( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \):

\( x^2 + \frac{1}{x^2} = \left(x - \frac{1}{x}\right)^2 + 2 \)

Substituting:

\( x^2 + \frac{1}{x^2} = (3)^2 + 2 = 9 + 2 = 7 \)

In simple words: Square the difference \( x - \frac{1}{x} \) and then add 2. This gives you the sum of x² and its reciprocal.

Exam Tip: Work with the reciprocal form directly - rearrange to get \( x - \frac{1}{x} \) first, then square it. Never try to find the actual value of x.

 

Question 11. If x + y = 8 and xy = 3¾, find the values of
(i) \( x - y \)
(ii) \( 3(x^2 + y^2) \)
(iii) \( 5(x^2 + y^2) + 4(x - y) \)
Answer:
(i) Using \( (x - y)^2 = (x + y)^2 - 4xy \):

\( (x - y)^2 = (8)^2 - 4 \times \frac{15}{4} = 64 - 15 = 49 \)

\( x - y = \pm 7 \)

(ii) Using \( x^2 + y^2 = (x + y)^2 - 2xy \):

\( x^2 + y^2 = (8)^2 - 2 \times \frac{15}{4} = 64 - \frac{15}{2} = \frac{128 - 15}{2} = \frac{113}{2} \)

\( 3(x^2 + y^2) = 3 \times \frac{113}{2} = \frac{339}{2} = 169\frac{1}{2} \)

(iii) When \( x - y = 7 \):

\( 5(x^2 + y^2) + 4(x - y) = 5 \times \frac{113}{2} + 4 \times 7 = \frac{565}{2} + 28 = \frac{565 + 56}{2} = \frac{621}{2} = 310\frac{1}{2} \)

When \( x - y = -7 \):

\( 5(x^2 + y^2) + 4(x - y) = 5 \times \frac{113}{2} + 4 \times (-7) = \frac{565}{2} - 28 = \frac{565 - 56}{2} = \frac{509}{2} = 254\frac{1}{2} \)

In simple words: Find x - y first using the formula. Then find x² + y² by rearranging the (x + y)² identity. Multiply by the needed coefficient and add the remaining term.

Exam Tip: Watch for fractional values of xy - use the correct form when substituting. Part (iii) depends on which ± value you choose in part (i), so show both cases.

 

Question 12. If x² + y² = 34 and xy = 10½, find the value of 2(x + y)² + (x - y)².
Answer: We begin by expanding the given expression. Using the algebraic identities \( 2(x + y)^2 + (x - y)^2 = 2(x^2 + y^2 + 2xy) + (x^2 + y^2 - 2xy) \). Simplifying: \( = 2x^2 + 2y^2 + 4xy + x^2 + y^2 - 2xy = 3x^2 + 3y^2 + 2xy = 3(x^2 + y^2) + 2xy \). Now we substitute the given values: \( = 3 \times 34 + 2 \times 10.5 = 102 + 21 = 123 \).
In simple words: First expand the expression using basic algebra rules. Then replace x² + y² with 34 and xy with 10.5. Do the arithmetic to get 123.

Exam Tip: Always expand both squared terms completely before substituting values - skipping this step often leads to careless errors.

 

Question 13. If a - b = 3 and ab = 4, find a³ - b³.
Answer: We use the algebraic identity for the difference of cubes. From \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we get \( a^3 - b^3 = (a - b)^3 + 3ab(a - b) \). Substituting the given values: \( a^3 - b^3 = (3)^3 + 3 \times 4 \times 3 = 27 + 36 = 63 \).
In simple words: Use the cube formula that connects a³ - b³ to (a - b)³. Plug in the numbers and add to get 63.

Exam Tip: Remember the identity \( a^3 - b^3 = (a - b)^3 + 3ab(a - b) \) - this is much faster than trying to compute a and b individually.

 

Question 14. If 2a - 3b = 3 and ab = 2, find the value of 8a³ - 27b³.
Answer: Notice that \( 8a^3 - 27b^3 = (2a)^3 - (3b)^3 \). Using the difference of cubes identity: \( (2a)^3 - (3b)^3 = (2a - 3b)^3 + 3 \times 2a \times 3b(2a - 3b) \). This gives us: \( 8a^3 - 27b^3 = (2a - 3b)^3 + 18ab(2a - 3b) \). Substituting the known values: \( = (3)^3 + 18 \times 2 \times 3 = 27 + 108 = 135 \).
In simple words: Rewrite 8a³ as (2a)³ and 27b³ as (3b)³. Then apply the cube difference formula with these new expressions and substitute to get 135.

Exam Tip: Always identify when expressions like 8a³ and 27b³ can be written as perfect cubes - recognizing this form is the key to solving such problems quickly.

 

Question 15. If x + 1/x = 4, find the values of
(i) x² + 1/x²
(ii) x⁴ + 1/x⁴
(iii) x³ + 1/x³
(iv) x - 1/x
Answer:
(i) Starting from the identity \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \), we find \( x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = (4)^2 - 2 = 16 - 2 = 14 \).
(ii) Using \( x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (14)^2 - 2 = 196 - 2 = 194 \).
(iii) Using the identity \( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = (4)^3 - 3 \times 4 = 64 - 12 = 52 \).
(iv) From \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \), we get \( x - \frac{1}{x} = \sqrt{14 - 2} = \sqrt{12} = 2\sqrt{3} \).
In simple words: These problems use the squaring and cubing identities. Once you find x² + 1/x², you can use it to find higher powers. Keep applying the formulas step by step.

Exam Tip: Build answers progressively - don't try to jump to x⁴ + 1/x⁴ without first finding x² + 1/x². Each part sets up the next part.

 

Question 16. If x - 1/x = 5, find the value of x⁴ + 1/x⁴.
Answer: First, we find \( x^2 + \frac{1}{x^2} \) using \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \), so \( x^2 + \frac{1}{x^2} = (5)^2 + 2 = 27 \). Next, we apply \( x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (27)^2 - 2 = 729 - 2 = 727 \).
In simple words: Square the first expression to get x² + 1/x². Then square that result and subtract 2 to get x⁴ + 1/x⁴.

Exam Tip: Always find the intermediate result (x² + 1/x²) first - trying to jump directly to x⁴ + 1/x⁴ almost always leads to mistakes.

 

Question 17. If x - 1/x = √5, find the values of
(i) x² + 1/x²
(ii) x + 1/x
(iii) x³ + 1/x³
Answer:
(i) From \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 \), we get \( x^2 + \frac{1}{x^2} = (\sqrt{5})^2 + 2 = 5 + 2 = 7 \).
(ii) From \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \), we find \( x + \frac{1}{x} = \sqrt{7 + 2} = \sqrt{9} = \pm 3 \).
(iii) Using \( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \): When \( x + \frac{1}{x} = 3 \), we get \( x^3 + \frac{1}{x^3} = (3)^3 - 3 \times 3 = 27 - 9 = 18 \). When \( x + \frac{1}{x} = -3 \), we get \( x^3 + \frac{1}{x^3} = (-3)^3 - 3 \times (-3) = -27 + 9 = -18 \).
In simple words: Find x² + 1/x² first. Use that to find x + 1/x, which comes out as either 3 or -3. Finally, use each value to calculate x³ + 1/x³, giving two possible answers.

Exam Tip: Watch for cases where you get two possible values - show both clearly and explain why each comes from a different value of x + 1/x.

 

Question 18. If x + 1/x = 6, find
(i) x - 1/x
(ii) x² - 1/x²
Answer:
(i) Using the relationship \( (x - \frac{1}{x})^2 = (x + \frac{1}{x})^2 - 4 \), we find \( x - \frac{1}{x} = \sqrt{(6)^2 - 4} = \sqrt{36 - 4} = \sqrt{32} = 4\sqrt{2} \).
(ii) Since \( x^2 - \frac{1}{x^2} = (x + \frac{1}{x})(x - \frac{1}{x}) \): When \( x - \frac{1}{x} = 4\sqrt{2} \), we get \( x^2 - \frac{1}{x^2} = 6 \times 4\sqrt{2} = 24\sqrt{2} \). When \( x - \frac{1}{x} = -4\sqrt{2} \), we get \( x^2 - \frac{1}{x^2} = 6 \times (-4\sqrt{2}) = -24\sqrt{2} \).
In simple words: Find x - 1/x using a square-root formula. Then multiply that by x + 1/x to get x² - 1/x², giving two possible answers depending on the sign.

Exam Tip: Remember the difference-of-squares factorization: x² - 1/x² = (x + 1/x)(x - 1/x) - this saves time and reduces errors.

 

Question 19. If x + 1/x = 2, prove that x² + 1/x² = x³ + 1/x³ = x⁴ + 1/x⁴.
Answer: Finding \( x^2 + \frac{1}{x^2} \): Using \( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \), we get \( x^2 + \frac{1}{x^2} = (2)^2 - 2 = 4 - 2 = 2 \) ... (i). Finding \( x^3 + \frac{1}{x^3} \): Using \( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \), we get \( x^3 + \frac{1}{x^3} = (2)^3 - 3 \times 2 = 8 - 6 = 2 \) ... (ii). Finding \( x^4 + \frac{1}{x^4} \): Using \( x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 \), we get \( x^4 + \frac{1}{x^4} = (2)^2 - 2 = 4 - 2 = 2 \) ... (iii). From (i), (ii), and (iii), all three expressions equal 2, hence proved.
In simple words: Apply the three different formulas for x² + 1/x², x³ + 1/x³, and x⁴ + 1/x⁴. When x + 1/x = 2, all three come out equal to 2 - a special case that shows up nicely here.

Exam Tip: This type of problem tests whether you know all three identities. Show your work clearly for each expression to earn full marks.

 

Question 20. If x - 2/x = 3, find the value of x³ - 8/x³.
Answer: We use the identity for the difference of cubes. From \( (a - b)^3 = a^3 - b^3 - 3ab(a - b) \), we rearrange to get \( a^3 - b^3 = (a - b)^3 + 3ab(a - b) \). Here, \( a = x \) and \( b = \frac{2}{x} \), so \( x^3 - \frac{8}{x^3} = (x - \frac{2}{x})^3 + 3 \times x \times \frac{2}{x}(x - \frac{2}{x}) = (x - \frac{2}{x})^3 + 6(x - \frac{2}{x}) \). Substituting \( x - \frac{2}{x} = 3 \): \( x^3 - \frac{8}{x^3} = (3)^3 + 6 \times 3 = 27 + 18 = 45 \).
In simple words: Recognize that 8/x³ = (2/x)³. Then use the difference of cubes formula with a = x and b = 2/x, plug in the value, and calculate to get 45.

Exam Tip: Always check whether an expression can be written as a difference of cubes - if it can, use the cube identity to simplify quickly.

 

Question 21. If a + 2b = 5, prove that a³ + 8b³ + 30ab = 125.
Answer: We expand using the cube formula: \( (a + 2b)^3 = a^3 + 8b^3 + 3(a)(2b)(a + 2b) \). This becomes \( (a + 2b)^3 = a^3 + 8b^3 + 6ab(a + 2b) \). Substituting \( a + 2b = 5 \) into both sides: \( (5)^3 = a^3 + 8b^3 + 6ab \times 5 \). Simplifying: \( 125 = a^3 + 8b^3 + 30ab \), hence proved.
In simple words: Use the cube expansion formula for (a + 2b)³. Since a + 2b is given as 5, cube it to get 125. Rearranging the formula gives the required result.

Exam Tip: Proof-type questions require you to show each step clearly - write out the full formula, substitute the known value, and rearrange to reach the target expression.

 

Question 22. If a + 1/a = p, prove that a³ + 1/a³ = p(p² - 3).
Answer: Using the cube formula: \( (a + \frac{1}{a})^3 = a^3 + \frac{1}{a^3} + 3(a + \frac{1}{a}) \). Rearranging: \( a^3 + \frac{1}{a^3} = (a + \frac{1}{a})^3 - 3(a + \frac{1}{a}) \). Substituting \( a + \frac{1}{a} = p \): \( a^3 + \frac{1}{a^3} = p^3 - 3p = p(p^2 - 3) \), hence proved.
In simple words: Start with the cube expansion formula. Replace a + 1/a with the variable p everywhere. Factor out p at the end to get the required form.

Exam Tip: Working with variables like p instead of numbers is the same process - apply the formula and factor the final answer to match the required format.

 

Question 24. If x² + 1/x² = 27, find the value of 3x³ + 5x - 3/x³ - 5/x.
Answer: First, we find \( x - \frac{1}{x} \) from \( (x - \frac{1}{x})^2 = x^2 + \frac{1}{x^2} - 2 = 27 - 2 = 25 \), so \( x - \frac{1}{x} = 5 \). Next, we find \( x^3 - \frac{1}{x^3} \) using \( x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x}) = (5)^3 + 3 \times 5 = 125 + 15 = 140 \). Now we separate the target expression: \( 3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3(x^3 - \frac{1}{x^3}) + 5(x - \frac{1}{x}) = 3 \times 140 + 5 \times 5 = 420 + 25 = 445 \).
In simple words: From x² + 1/x², find x - 1/x first. Then use that to find x³ - 1/x³. Finally, regroup the target expression as 3(x³ - 1/x³) + 5(x - 1/x) and substitute the values you found.

Exam Tip: When an expression mixes x³ and x terms, try grouping them as separate blocks - this makes substitution straightforward and reduces errors.

 

Question 24. Simplify \( 3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} \) when \( x - \frac{1}{x} = 5 \).
Answer: The expression can be rewritten as \( 3\left(x^3 - \frac{1}{x^3}\right) + 5\left(x - \frac{1}{x}\right) \).

Using the identity \( x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \), we get:
\[ x^3 - \frac{1}{x^3} = (5)^3 + 3(5) = 125 + 15 = 140 \]

So, \( 3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3(140) + 5(5) = 420 + 25 = 445 \).

When \( x - \frac{1}{x} = -5 \):
\[ x^3 - \frac{1}{x^3} = (-5)^3 + 3(-5) = -125 - 15 = -140 \]
So, \( 3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = 3(-140) + 5(-5) = -420 - 25 = -445 \).

Therefore, \( 3x^3 + 5x - \frac{3}{x^3} - \frac{5}{x} = \pm 445 \).
In simple words: Break the expression into two parts using the cube difference formula. Substitute the given value to find the answer.

Exam Tip: Recognize the factorization pattern \( 3\left(x^3 - \frac{1}{x^3}\right) + 5\left(x - \frac{1}{x}\right) \) to use the cube identity efficiently. Check both cases for \( x - \frac{1}{x} = 5 \) and \( x - \frac{1}{x} = -5 \).

 

Question 25. If \( x^2 + \frac{1}{25x^2} = 8\frac{3}{5} \), find \( x + \frac{1}{5x} \).
Answer: We know that \( x^2 + \frac{1}{25x^2} = \left(x\right)^2 + \left(\frac{1}{5x}\right)^2 \).

Using the identity \( \left(x + \frac{1}{5x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{5x} + \frac{1}{25x^2} = x^2 + \frac{2}{5} + \frac{1}{25x^2} \), we get:
\[ x^2 + \frac{1}{25x^2} = \left(x + \frac{1}{5x}\right)^2 - \frac{2}{5} \]

Let \( x + \frac{1}{5x} = a \). Then:
\[ 8\frac{3}{5} = a^2 - \frac{2}{5} \]
\[ \frac{43}{5} = \frac{5a^2 - 2}{5} \]
\[ 5a^2 - 2 = 43 \]
\[ 5a^2 = 45 \]
\[ a^2 = 9 \]
\[ a = \pm 3 \]

Therefore, \( x + \frac{1}{5x} = \pm 3 \).
In simple words: Square the expression you want to find, relate it to the given information, and solve for the unknown.

Exam Tip: Watch for perfect square patterns when dealing with sum of squares expressions. The constant term \( \frac{2}{5} \) is crucial—forgetting it is a common error.

 

Question 26. If \( x^2 + \frac{1}{4x^2} = 8 \), find \( x^3 + \frac{1}{8x^3} \).
Answer: We know that \( \left(x + \frac{1}{2x}\right)^2 = x^2 + \frac{1}{4x^2} + 2 \cdot x \cdot \frac{1}{2x} = x^2 + \frac{1}{4x^2} + 1 \).

From the given information:
\[ x^2 + \frac{1}{4x^2} = \left(x + \frac{1}{2x}\right)^2 - 1 \]
\[ 8 = \left(x + \frac{1}{2x}\right)^2 - 1 \]
\[ \left(x + \frac{1}{2x}\right)^2 = 9 \]
\[ x + \frac{1}{2x} = \pm 3 \]

Using the identity \( x^3 + \frac{1}{8x^3} = \left(x + \frac{1}{2x}\right)^3 - 3\left(x + \frac{1}{2x}\right) \):

When \( x + \frac{1}{2x} = 3 \):
\[ x^3 + \frac{1}{8x^3} = (3)^3 - 3(3) = 27 - 9 = 18 \]

When \( x + \frac{1}{2x} = -3 \):
\[ x^3 + \frac{1}{8x^3} = (-3)^3 - 3(-3) = -27 + 9 = -18 \]

Therefore, \( x^3 + \frac{1}{8x^3} = \pm 18 \).
In simple words: First, find \( x + \frac{1}{2x} \) from the given square expression. Then use the cube formula to get the final answer.

Exam Tip: Apply the square-to-cube progression: first solve for the linear sum, then use it to find the cubic sum using the appropriate identity.

 

Question 27. If \( a^2 - 3a + 1 = 0 \), find
(i) \( a^2 + \frac{1}{a^2} \)
(ii) \( a^3 + \frac{1}{a^3} \)
Answer: Dividing each term of \( a^2 - 3a + 1 = 0 \) by \( a \):
\[ a + \frac{1}{a} = 3 \]

(i) Using the identity \( a^2 + \frac{1}{a^2} = \left(a + \frac{1}{a}\right)^2 - 2 \):
\[ a^2 + \frac{1}{a^2} = (3)^2 - 2 = 9 - 2 = 7 \]

(ii) Using the identity \( a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)^3 - 3\left(a + \frac{1}{a}\right) \):
\[ a^3 + \frac{1}{a^3} = (3)^3 - 3(3) = 27 - 9 = 18 \]
In simple words: Divide the given equation to find \( a + \frac{1}{a} \). Use this result with standard formulas to find squares and cubes.

Exam Tip: Always divide by the variable first when given a quadratic equation relating a variable and its reciprocal—this creates a simpler working equation.

 

Question 28. If \( a = \frac{1}{a - 5} \), find
(i) \( a - \frac{1}{a} \)
(ii) \( a + \frac{1}{a} \)
(iii) \( a^2 - \frac{1}{a^2} \)
Answer: Given \( a = \frac{1}{a - 5} \), we get \( a(a - 5) = 1 \), which gives \( a^2 - 5a - 1 = 0 \).

(i) Dividing by \( a \):
\[ a - 5 - \frac{1}{a} = 0 \]
\[ a - \frac{1}{a} = 5 \]

(ii) We use the identity \( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \) and \( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \).

Subtracting these equations:
\[ \left(a + \frac{1}{a}\right)^2 - \left(a - \frac{1}{a}\right)^2 = 4 \]
\[ \left(a + \frac{1}{a}\right)^2 = \left(a - \frac{1}{a}\right)^2 + 4 = (5)^2 + 4 = 29 \]
\[ a + \frac{1}{a} = \pm\sqrt{29} \]

(iii) Using the identity \( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \):
\[ a^2 - \frac{1}{a^2} = \pm\sqrt{29} \times 5 = \pm 5\sqrt{29} \]
In simple words: From the given condition, derive a quadratic equation. Use the difference and sum identities to connect the unknowns step by step.

Exam Tip: When you have both sum and difference expressions, subtract their squares to find a useful relationship. Remember to consider both positive and negative roots.

 

Question 29. If \( \left(x + \frac{1}{x}\right)^2 = 3 \), find \( x^3 + \frac{1}{x^3} \).
Answer: From the given condition:
\[ x + \frac{1}{x} = \pm\sqrt{3} \]

Using the identity \( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \):

When \( x + \frac{1}{x} = \sqrt{3} \):
\[ x^3 + \frac{1}{x^3} = (\sqrt{3})^3 - 3\sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0 \]

When \( x + \frac{1}{x} = -\sqrt{3} \):
\[ x^3 + \frac{1}{x^3} = (-\sqrt{3})^3 - 3(-\sqrt{3}) = -3\sqrt{3} + 3\sqrt{3} = 0 \]

Therefore, \( x^3 + \frac{1}{x^3} = 0 \).
In simple words: Take the square root to find \( x + \frac{1}{x} \). Then use the cube formula, which simplifies nicely to give zero in both cases.

Exam Tip: When a cube formula yields terms that cancel, check your work carefully—this is often a sign you are on the right track, not that you made an error.

 

Question 30. If \( x = 5 - 2\sqrt{6} \), find the value of \( \sqrt{x} + \frac{1}{\sqrt{x}} \).
Answer: First, find \( \frac{1}{x} \) by rationalizing:
\[ \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} \cdot \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6}}{25 - 24} = 5 + 2\sqrt{6} \]

Therefore:
\[ x + \frac{1}{x} = 5 - 2\sqrt{6} + 5 + 2\sqrt{6} = 10 \]

Using the identity \( \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = x + \frac{1}{x} + 2 \):
\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = 10 + 2 = 12 \]
\[ \sqrt{x} + \frac{1}{\sqrt{x}} = \pm 2\sqrt{3} \]
In simple words: Simplify \( x + \frac{1}{x} \) by rationalizing the reciprocal. Then square the expression you seek and take its square root.

Exam Tip: Rationalization is essential when dealing with surds. Ensure the product of conjugates simplifies to an integer before moving forward.

 

Question 31. If \( a + b + c = 12 \) and \( ab + bc + ca = 22 \), find \( a^2 + b^2 + c^2 \).
Answer: Using the identity \( a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) \):
\[ a^2 + b^2 + c^2 = (12)^2 - 2(22) = 144 - 44 = 100 \]
In simple words: Square the sum of the three numbers. Then subtract twice the sum of their pairwise products to get the sum of their squares.

Exam Tip: This is a fundamental identity—memorizing and applying it directly saves time on algebraic expansion.

 

Question 32. If \( a + b + c = 12 \) and \( a^2 + b^2 + c^2 = 100 \), find \( ab + bc + ca \).
Answer: Using the identity \( a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) \):
\[ 100 = (12)^2 - 2(ab + bc + ca) \]
\[ 100 = 144 - 2(ab + bc + ca) \]
\[ 2(ab + bc + ca) = 144 - 100 = 44 \]
\[ ab + bc + ca = 22 \]
In simple words: Rearrange the identity to solve for the product sum when you know the other two quantities.

Exam Tip: This is the inverse operation of Question 31—rearrange the formula algebraically to isolate the term you need.

 

Question 33. If \( a^2 + b^2 + c^2 = 125 \) and \( ab + bc + ca = 50 \), find \( a + b + c \).
Answer: Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \):
\[ (a + b + c)^2 = 125 + 2(50) = 125 + 100 = 225 \]
\[ a + b + c = \pm 15 \]
In simple words: Add twice the pairwise products to the sum of squares, then take the square root of the result.

Exam Tip: Remember to include both positive and negative square roots unless the problem context rules out one solution.

 

Question 34. If \( a + b - c = 5 \) and \( a^2 + b^2 + c^2 = 29 \), find the value of \( ab - bc - ca \).
Answer: Squaring both sides of \( a + b - c = 5 \):
\[ (a + b - c)^2 = 25 \]
\[ a^2 + b^2 + c^2 + 2ab - 2bc - 2ca = 25 \]

Substituting \( a^2 + b^2 + c^2 = 29 \):
\[ 29 + 2(ab - bc - ca) = 25 \]
\[ 2(ab - bc - ca) = 25 - 29 = -4 \]
\[ ab - bc - ca = -2 \]
In simple words: Square the given linear constraint to create an equation involving the squares. Substitute the known sum of squares and solve for the product combination.

Exam Tip: Watch the sign patterns when expanding \( (a + b - c)^2 \)—the negative term distributes with opposite signs to some products.

 

Question 35. If \( a - b = 7 \) and \( a^2 + b^2 = 85 \), then find the value of \( a^3 - b^3 \).
Answer: Using the identity \( (a - b)^2 = a^2 + b^2 - 2ab \):
\[ (7)^2 = 85 - 2ab \]
\[ 49 = 85 - 2ab \]
\[ 2ab = 36 \]
\[ ab = 18 \]

Using the identity \( a^3 - b^3 = (a - b)(a^2 + b^2 + ab) \):
\[ a^3 - b^3 = 7(85 + 18) = 7(103) = 721 \]
In simple words: Find the product \( ab \) from the difference and sum of squares. Then use the factorization of difference of cubes.

Exam Tip: The difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) is more practical than direct expansion when you know these intermediate values.

 

Question 36. If the number \( x \) is 3 less than the number \( y \) and the sum of the squares of \( x \) and \( y \) is 29, find the product of \( x \) and \( y \).
Answer: Given:
\[ x = y - 3 \text{ or } x - y = -3 \]
\[ x^2 + y^2 = 29 \]

Using the identity \( (x + y)^2 = x^2 + y^2 + 2xy \) and \( (x - y)^2 = x^2 + y^2 - 2xy \):
\[ (-3)^2 = 29 - 2xy \]
\[ 9 = 29 - 2xy \]
\[ 2xy = 20 \]
\[ xy = 10 \]
In simple words: Use the constraint \( x - y = -3 \) along with the sum of squares to find the product through the difference-of-squares identity.

Exam Tip: When given a linear relationship and a sum of squares, set up both sum and difference identities—one will directly give you the product without needing to find individual values.

 

Question 37. If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.
Answer: Let the two numbers be x and y. We are given that x + y = 8 and xy = 15. Using the identity x³ + y³ = (x + y)³ - 3xy(x + y), we get x³ + y³ = 8³ - 3(15)(8) = 512 - 360 = 152.
In simple words: When you know the sum and product of two numbers, you can find the sum of their cubes by plugging these values into a formula. Here, the answer comes out to 152.

Exam Tip: Remember the identity x³ + y³ = (x + y)³ - 3xy(x + y) and substitute the given values carefully to avoid arithmetic errors.

 

Multiple Choice Questions

 

Question 1. If \( x + \frac{1}{x} = 2 \), then \( x^2 + \frac{1}{x^2} \) is equal to
(a) 4
(b) 2
(c) 0
(d) none of these
Answer: (b) 2
In simple words: When you square the expression \( x + \frac{1}{x} = 2 \) and simplify, the middle term cancels out, leaving \( x^2 + \frac{1}{x^2} = 2 \).

Exam Tip: Use the identity \( (a + b)^2 = a^2 + b^2 + 2ab \) and rearrange to isolate the required term.

 

Question 2. If \( x^2 + y^2 = 9 \) and \( xy = 8 \), then \( x + y \) is equal to
(a) 25
(b) 5
(c) -5
(d) ±5
Answer: (d) ±5
In simple words: The identity \( (x + y)^2 = x^2 + y^2 + 2xy \) gives us \( (x + y)^2 = 9 + 16 = 25 \), so \( x + y = ±5 \).

Exam Tip: When taking the square root, remember that both positive and negative values are possible solutions unless the problem specifies otherwise.

 

Question 3. \( (102)^2 - (98)^2 \) is equal to
(a) 200
(b) 400
(c) 600
(d) 800
Answer: (d) 800
In simple words: Using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \), we calculate (102 - 98)(102 + 98) = 4 × 200 = 800.

Exam Tip: The difference of squares formula is one of the quickest methods for this type of calculation - always check if numbers are close together.

 

Question 4. \( 96 × 104 \) is equal to
(a) 9984
(b) 9974
(c) 9964
(d) none of these
Answer: (a) 9984
In simple words: Rewrite 96 × 104 as (100 - 4)(100 + 4). By the difference of squares formula, this equals 100² - 4² = 10000 - 16 = 9984.

Exam Tip: Look for numbers that are symmetrically placed around a convenient base (like 100) to apply the difference of squares formula efficiently.

 

Question 5. \( \frac{103^2 - 97^2}{200} \) is equal to
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (d) 6
In simple words: Apply the difference of squares: 103² - 97² = (103 - 97)(103 + 97) = 6 × 200. Dividing by 200 gives 6.

Exam Tip: Always check if the denominator can be simplified by factoring the numerator - it often cancels neatly.

 

Question 6. If \( x + y = 11 \) and \( xy = 24 \), then \( x^2 + y^2 \) is equal to
(a) 121
(b) 73
(c) 48
(d) 169
Answer: (b) 73
In simple words: Use the identity \( x^2 + y^2 = (x + y)^2 - 2xy \). Substituting gives 11² - 2(24) = 121 - 48 = 73.

Exam Tip: This identity \( x^2 + y^2 = (x + y)^2 - 2xy \) is fundamental and appears frequently - commit it to memory.

 

Question 7. The value of \( 249^2 - 248^2 \) is
(a) 12
(b) 477
(c) 487
(d) 497
Answer: (d) 497
In simple words: Using \( a^2 - b^2 = (a - b)(a + b) \), we get (249 - 248)(249 + 248) = 1 × 497 = 497.

Exam Tip: When the two numbers differ by just 1, the difference of their squares equals their sum - a useful shortcut.

 

Question 8. If \( \frac{x}{y} + \frac{y}{x} = -1 \) (x, y ≠ 0), then the value of \( x^3 - y^3 \) is
(a) 1
(b) -1
(c) 0
(d) \( \frac{1}{2} \)
Answer: (c) 0
In simple words: From the given condition, we can show that \( x^2 + y^2 + xy = 0 \). Using the factorization \( x^3 - y^3 = (x - y)(x^2 + y^2 + xy) \), we get (x - y) × 0 = 0.

Exam Tip: When you derive a key relationship from the given condition, check if it appears in any factorization formula you know - this often unlocks the solution quickly.

 

Question 9. If \( a + b + c = 0 \), then the value of \( a^3 + b^3 + c^3 \) is
(a) 0
(b) abc
(c) 2abc
(d) 3abc
Answer: (d) 3abc
In simple words: The identity \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - bc - ca - ab) \) shows that when a + b + c = 0, the right side becomes 0, leaving a³ + b³ + c³ = 3abc.

Exam Tip: This is a classical identity - whenever you see a + b + c = 0, think immediately of this relationship between a³ + b³ + c³ and abc.

 

Question 10. If \( x - \frac{2}{x} = 3 \), then \( x^3 - \frac{8}{x^3} \) is equal to
(a) 27
(b) 36
(c) 45
(d) 54
Answer: (c) 45
In simple words: Use the identity \( a^3 - b^3 = (a - b)^3 + 3ab(a - b) \). Substituting a = x and b = \( \frac{2}{x} \) with the given value gives 3³ + 6 × 3 = 27 + 18 = 45.

Exam Tip: For expressions involving \( a - b \) and \( a^3 - b^3 \), use the cubic difference identity rather than trying to compute the cube directly.

 

Question 11. Consider the following two statements:
Statement 1: If \( a + b = 0 \), then \( a^2 + b^2 = 0 \)
Statement 2: \( a^2 + b^2 = (a + b)^2 \)
Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (b) Both the statements are false.
In simple words: For Statement 1: if a + b = 0, then squaring both sides gives a² + 2ab + b² = 0, so a² + b² = -2ab, which is zero only if ab = 0 (not always true). For Statement 2: expanding (a + b)² gives a² + 2ab + b², which is not equal to a² + b² unless ab = 0.

Exam Tip: Always verify general algebraic statements with specific examples before concluding they are true - this catches mistakes quickly.

 

Assertion Reason Type Questions

 

Question. Assertion (A): If \( x = \frac{1}{x} \), then \( x = ± 1 \).
Reason (R): \( (x - \frac{1}{x})(x + \frac{1}{x}) = x^2 - \frac{1}{x^2} \)
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
In simple words: From x = 1/x, we get x² = 1, so x = ±1 (Assertion is true). The Reason states a valid algebraic identity (difference of squares), but this identity doesn't directly explain why x = ±1 from the assertion condition.

Exam Tip: In Assertion-Reason questions, verify both statements independently, then check if the reason logically supports the assertion - both can be true but unrelated.

 

Question. Assertion (A): 1003 × 997 = 999991
Reason (R): (a - b)(a + b) = a² - b²
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: We can rewrite 1003 × 997 as (1000 + 3)(1000 - 3). By the difference of squares formula, this equals 1000² - 3² = 1000000 - 9 = 999991. The Reason directly explains the Assertion.

Exam Tip: Recognize when two numbers can be expressed as (a + b) and (a - b) for some convenient base a - this setup always invites the difference of squares formula.

 

Chapter Test

 

Question 1. Find the expansions of the following:
(i) \( (2x + 3y + 5)(2x + 3y - 5) \)
(ii) \( (6 - 4a - 7b)^2 \)
(iii) \( (7 - 3xy)^3 \)
(iv) \( (x + y + 2)^3 \)
Answer:
(i) Treating (2x + 3y) as a single block, we recognize this as a difference of squares: \( (2x + 3y + 5)(2x + 3y - 5) = (2x + 3y)^2 - 5^2 = 4x^2 + 9y^2 + 12xy - 25 \)
(ii) Using the identity for a trinomial square: \( (6 - 4a - 7b)^2 = (6 - 4a)^2 + (7b)^2 - 2(6 - 4a)(7b) = 36 + 16a^2 + 49b^2 - 48a + 56ab - 84b \)
(iii) Applying the cube difference formula: \( (7 - 3xy)^3 = 7^3 - (3xy)^3 - 3(7)(3xy)(7 - 3xy) = 343 - 27x^3y^3 - 441xy + 189x^2y^2 \)
(iv) Grouping (y + 2) and treating it as a unit: \( (x + y + 2)^3 = x^3 + (y + 2)^3 + 3x(y + 2)(x + y + 2) \). This expands fully to give multiple terms involving x, y, and their powers.
In simple words: Each of these expressions uses standard algebraic identities - difference of squares for (i), trinomial square for (ii), and cubic formulas for (iii) and (iv). Apply the appropriate identity to expand each one.

Exam Tip: Always identify the pattern in an expression before expanding - recognizing when a difference of squares or cube identity applies saves time and reduces errors.

 

Question 1. Expand \( (x + y + 2)^3 \) and simplify.
Answer: Using the expansion formula, we can write \( (x + y + 2)^3 = x^3 + y^3 + 8 + 3(y)(2)(y + 2) + (3xy + 6x)(x + y + 2) \). Expanding further:
\( (x + y + 2)^3 = x^3 + y^3 + 8 + 6y^2 + 12y + 3x^2y + 3xy^2 + 6xy + 6x^2 + 6xy + 12x \)
Collecting like terms together, the result is:
\( (x + y + 2)^3 = x^3 + y^3 + 3x^2y + 3xy^2 + 6x^2 + 6y^2 + 12xy + 12x + 12y + 8 \)
In simple words: When you cube the sum of three terms, expand each part using the binomial rule and then gather all terms with the same power together.

Exam Tip: Always arrange the final expanded form in descending order of powers and double-check by counting how many distinct terms you should have before writing the answer.

 

Question 2. Simplify \( (x - 2)(x + 2)(x^2 + 4)(x^4 + 16) \).
Answer: Apply the difference-of-squares formula \( (a - b)(a + b) = a^2 - b^2 \) repeatedly:
\( (x - 2)(x + 2) = x^2 - 4 \)
\( (x^2 - 4)(x^2 + 4) = x^4 - 16 \)
\( (x^4 - 16)(x^4 + 16) = (x^4)^2 - 16^2 = x^8 - 256 \)
In simple words: When you see pairs that look like \( (a - b) \) and \( (a + b) \), multiply them together to get \( a^2 - b^2 \), then repeat the pattern.

Exam Tip: Identify the difference-of-squares pattern at each step - this method is much faster than expanding everything at once.

 

Question 3. Evaluate \( 1002 \times 998 \) by using a special product.
Answer: Express the numbers as \( 1002 \times 998 = (1000 + 2)(1000 - 2) \). Using the difference-of-squares identity \( (a - b)(a + b) = a^2 - b^2 \):
\( (1000 + 2)(1000 - 2) = 1000^2 - 2^2 = 1,000,000 - 4 = 999,996 \)
In simple words: Rewrite numbers close to a round value as that round value plus or minus a small difference, then apply the difference-of-squares rule to get the answer quickly.

Exam Tip: This technique avoids long multiplication - look for patterns where two numbers are equidistant from a central round number.

 

Question 4. If \( a + 2b + 3c = 0 \), prove that \( a^3 + 8b^3 + 27c^3 = 18abc \).
Answer: From the given condition, rearrange to get \( a + 2b = -3c \). Cube both sides:
\( (a + 2b)^3 = (-3c)^3 \)
Expanding the left side using the sum-of-cubes expansion:
\( a^3 + (2b)^3 + 3(a)(2b)(a + 2b) = -27c^3 \)
\( a^3 + 8b^3 + 6ab(-3c) = -27c^3 \)
\( a^3 + 8b^3 - 18abc = -27c^3 \)
Moving \( 27c^3 \) to the left side gives the required result:
\( a^3 + 8b^3 + 27c^3 = 18abc \)
In simple words: Use the given constraint to replace one part of the expression, then cube and expand carefully to reach the target form.

Exam Tip: When a constraint is given as a sum equal to zero, always rearrange it to isolate one variable in terms of the others before cubing.

 

Question 5. If \( 2x = 3y - 5 \), then find the value of \( 8x^3 - 27y^3 - 90xy + 125 \).
Answer: Rearrange the given condition as \( 2x - 3y = -5 \). Cube both sides:
\( (2x - 3y)^3 = (-5)^3 \)
\( (2x)^3 - (3y)^3 - 3(2x)(3y)(2x - 3y) = -125 \)
\( 8x^3 - 27y^3 - 18xy(2x - 3y) = -125 \)
Substitute \( 2x - 3y = -5 \):
\( 8x^3 - 27y^3 - 18xy(-5) = -125 \)
\( 8x^3 - 27y^3 + 90xy = -125 \)
\( 8x^3 - 27y^3 + 90xy + 125 = 0 \)
In simple words: Rearrange the given condition, cube it to create the expression you need, and then substitute back to find the answer.

Exam Tip: Always cube the entire constraint equation, not just individual terms, to maintain the relationship needed for substitution.

 

Question 6. If \( a^2 - \frac{1}{a^2} = 5 \), evaluate \( a^4 + \frac{1}{a^4} \).
Answer: Use the algebraic identity:
\( a^4 + \frac{1}{a^4} = \left( a^2 - \frac{1}{a^2} \right)^2 + 2 \)
Substitute the given value:
\( a^4 + \frac{1}{a^4} = (5)^2 + 2 = 25 + 2 = 27 \)
In simple words: Square the given expression and add 2 to get the fourth power sum.

Exam Tip: Remember the identity \( \left( a^2 - \frac{1}{a^2} \right)^2 + 2 = a^4 + \frac{1}{a^4} \) - it's faster than squaring manually.

 

Question 7. If \( a + \frac{1}{a} = p \) and \( a - \frac{1}{a} = q \), find the relation between p and q.
Answer: Square both expressions:
\( \left( a + \frac{1}{a} \right)^2 = a^2 + \frac{1}{a^2} + 2 = p^2 \) ... (i)
\( \left( a - \frac{1}{a} \right)^2 = a^2 + \frac{1}{a^2} - 2 = q^2 \) ... (ii)
Subtract equation (ii) from equation (i):
\( \left( a + \frac{1}{a} \right)^2 - \left( a - \frac{1}{a} \right)^2 = 4 \)
\( p^2 - q^2 = 4 \)
In simple words: Square each expression separately, subtract one from the other, and the middle terms cancel to leave a simple relation between p and q.

Exam Tip: When two expressions are defined, always try squaring them individually first - the resulting equations often contain useful common terms that cancel.

 

Question 8. If \( \frac{a^2 + 1}{a} = 4 \), find the value of \( 2a^3 + \frac{2}{a^3} \).
Answer: From the given condition, divide both sides by a:
\( a + \frac{1}{a} = 4 \)
To find the cube sum, use the identity:
\( a^3 + \frac{1}{a^3} = \left( a + \frac{1}{a} \right)^3 - 3 \left( a + \frac{1}{a} \right) \)
\( a^3 + \frac{1}{a^3} = (4)^3 - 3(4) = 64 - 12 = 52 \)
Therefore:
\( 2a^3 + \frac{2}{a^3} = 2 \left( a^3 + \frac{1}{a^3} \right) = 2(52) = 104 \)
In simple words: First simplify the given condition, then apply the cube-sum identity, and finally multiply by 2.

Exam Tip: Learn and use the identity for \( a^3 + \frac{1}{a^3} \) in terms of \( a + \frac{1}{a} \) - it appears frequently in algebra problems.

 

Question 9. If \( x = \frac{1}{4 - x} \), find the values of
(i) \( x + \frac{1}{x} \)
(ii) \( x^3 + \frac{1}{x^3} \)
(iii) \( x^6 + \frac{1}{x^6} \)
Answer:
(i) From the given condition, multiply both sides by \( (4 - x) \):
\( x(4 - x) = 1 \)
\( 4x - x^2 = 1 \)
Divide the entire equation by x:
\( 4 - x = \frac{1}{x} \)
\( x + \frac{1}{x} = 4 \)
(ii) Use the identity:
\( x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \)
\( x^3 + \frac{1}{x^3} = (4)^3 - 3(4) = 64 - 12 = 52 \)
(iii) Use the identity:
\( x^6 + \frac{1}{x^6} = \left( x^3 + \frac{1}{x^3} \right)^2 - 2 \)
\( x^6 + \frac{1}{x^6} = (52)^2 - 2 = 2704 - 2 = 2702 \)
In simple words: Find the basic sum \( x + \frac{1}{x} \) first, then use identities to build up to higher powers.

Exam Tip: Always solve for \( x + \frac{1}{x} \) early - this is the foundation for finding all higher power sums using the standard identities.

 

Question 10. If \( x - \frac{1}{x} = 3 + 2\sqrt{2} \), find the value of \( \frac{1}{4} \left( x^3 - \frac{1}{x^3} \right) \).
Answer: Use the identity:
\( x^3 - \frac{1}{x^3} = \left( x - \frac{1}{x} \right)^3 + 3 \left( x - \frac{1}{x} \right) \)
First, evaluate \( \left( x - \frac{1}{x} \right)^3 \) where \( x - \frac{1}{x} = 3 + 2\sqrt{2} \):
\( \left( 3 + 2\sqrt{2} \right)^3 = (3)^3 + (2\sqrt{2})^3 + 3(3)(2\sqrt{2})(3 + 2\sqrt{2}) + \text{other terms} \)
After expanding and simplifying:
\( (3 + 2\sqrt{2})^3 = 27 + 16\sqrt{2} + 18\sqrt{2}(3 + 2\sqrt{2}) + 9 + 6\sqrt{2} \)
\( = 27 + 9 + 16\sqrt{2} + 54\sqrt{2} + 72 + 6\sqrt{2} = 108 + 76\sqrt{2} \)
Also, \( 3(x - \frac{1}{x}) = 3(3 + 2\sqrt{2}) = 9 + 6\sqrt{2} \)
Therefore:
\( x^3 - \frac{1}{x^3} = 108 + 76\sqrt{2} + 9 + 6\sqrt{2} = 117 + 82\sqrt{2} \)
Wait, let me recalculate. Using a cleaner approach:
\( x^3 - \frac{1}{x^3} = (3 + 2\sqrt{2})^3 + 3(3 + 2\sqrt{2}) = 108 + 76\sqrt{2} \)
\( \frac{1}{4} \left( x^3 - \frac{1}{x^3} \right) = \frac{1}{4} (108 + 76\sqrt{2}) = 27 + 19\sqrt{2} \)
In simple words: Expand the difference-of-cubes formula, substitute the given value, calculate the cube carefully, and divide the final result by 4.

Exam Tip: When dealing with surds in the base expression, expand the cube step-by-step and group rational and irrational terms separately to avoid errors.

 

Question 11. If \( x + \frac{1}{x} = 3\frac{1}{3} \), find the value of \( x^3 - \frac{1}{x^3} \).
Answer: Convert the mixed number: \( x + \frac{1}{x} = \frac{10}{3} \)
Use the identity:
\( x - \frac{1}{x} = \sqrt{ \left( x + \frac{1}{x} \right)^2 - 4 } \)
\( x - \frac{1}{x} = \sqrt{ \left( \frac{10}{3} \right)^2 - 4 } = \sqrt{ \frac{100}{9} - 4 } = \sqrt{ \frac{100 - 36}{9} } = \sqrt{ \frac{64}{9} } = \frac{8}{3} \)
Now use the difference-of-cubes identity:
\( x^3 - \frac{1}{x^3} = \left( x - \frac{1}{x} \right)^3 + 3 \left( x - \frac{1}{x} \right) \)
\( x^3 - \frac{1}{x^3} = \left( \frac{8}{3} \right)^3 + 3 \left( \frac{8}{3} \right) = \frac{512}{27} + 8 = \frac{512 + 216}{27} = \frac{728}{27} = 26\frac{26}{27} \)
In simple words: First find \( x - \frac{1}{x} \) using the relation between sum and difference, then apply the cube formula.

Exam Tip: Always check whether you need the sum or the difference form of the reciprocal identity before expanding to cubes.

 

Question 12. If \( x = 2 - \sqrt{3} \), then find the value of \( x^3 - \frac{1}{x^3} \).
Answer: First, find the reciprocal of x by rationalizing:
\( \frac{1}{x} = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3} \)
Calculate the difference:
\( x - \frac{1}{x} = (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3} \)
Cube both sides:
\( \left( x - \frac{1}{x} \right)^3 = (-2\sqrt{3})^3 = -24\sqrt{3} \)
Using the identity \( x^3 - \frac{1}{x^3} - 3 \left( x - \frac{1}{x} \right) = \left( x - \frac{1}{x} \right)^3 \):
\( x^3 - \frac{1}{x^3} = (-2\sqrt{3})^3 + 3(-2\sqrt{3}) = -24\sqrt{3} - 6\sqrt{3} = -30\sqrt{3} \)
In simple words: Rationalize the reciprocal, find the difference, cube it, and then use the expansion formula to find the answer.

Exam Tip: When x involves a surd, always rationalize its reciprocal first before attempting any further calculations.

 

Question 13. If the sum of two numbers is 7 and sum of their cubes is 133, find the sum of their squares.
Answer: Let the two numbers be x and y.
Given: \( x + y = 7 \) and \( x^3 + y^3 = 133 \)
Use the sum-of-cubes identity:
\( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \)
Substitute the known values:
\( 7^3 = 133 + 3xy \times 7 \)
\( 343 = 133 + 21xy \)
\( 21xy = 210 \)
\( xy = 10 \)
Now use the sum-of-squares identity:
\( (x + y)^2 = x^2 + y^2 + 2xy \)
\( 49 = x^2 + y^2 + 20 \)
\( x^2 + y^2 = 29 \)
In simple words: Use the sum-of-cubes formula to find the product, then apply the sum-of-squares formula to get the final answer.

Exam Tip: When both the sum and sum of cubes are given, always extract the product first using the cubic identity - the product is then used in the quadratic identity.

 

Question 14. If \( a - b = 7 \) and \( a^3 - b^3 = 133 \), find
(i) ab
(ii) \( a^2 + b^2 \)
Answer:
(i) Cube the difference equation:
\( (a - b)^3 = 7^3 \)
\( a^3 - b^3 - 3ab(a - b) = 343 \)
\( 133 - 3ab(7) = 343 \)
\( 133 - 21ab = 343 \)
\( -21ab = 210 \)
\( ab = -10 \)
(ii) Use the sum-of-squares identity in terms of difference and product:
\( a^2 + b^2 = (a - b)^2 + 2ab \)
\( a^2 + b^2 = 7^2 + 2(-10) = 49 - 20 = 29 \)
In simple words: Cube the given difference to extract the product, then use the product to calculate the sum of squares.

Exam Tip: For difference-based problems, cube the difference equation first to isolate the product term - this is faster than other approaches.

 

Question 15. Find the coefficient of \( x^2 \) in the expansion of \( (x^2 + x + 1)^2 + (x^2 - x + 1)^2 \).
Answer: Recognize that the expression can be simplified using the identity \( (p + q)^2 + (p - q)^2 = 2(p^2 + q^2) \):
Let \( p = x^2 + 1 \) and \( q = x \). Then:
\( (x^2 + x + 1)^2 + (x^2 - x + 1)^2 = [(x^2 + 1) + x]^2 + [(x^2 + 1) - x]^2 = 2[(x^2 + 1)^2 + x^2] \)
\( = 2[(x^2)^2 + 1 + 2x^2 + x^2] = 2[x^4 + 3x^2 + 1] = 2x^4 + 6x^2 + 2 \)
The coefficient of \( x^2 \) is \( \boxed{6} \)
In simple words: Notice that both expressions have a similar form, use the sum-of-squares identity to combine them, then expand and identify the \( x^2 \) term.

Exam Tip: Look for patterns in polynomial expressions - recognizing symmetric forms like \( (p + q)^2 + (p - q)^2 \) can significantly simplify expansion.

 

Question 16. If \( x^2 + y^2 + z^2 = xy + yz + zx \), prove that \( x = y = z \).
Answer: Rearrange the given equation:
\( x^2 + y^2 + z^2 - xy - yz - zx = 0 \)
Multiply both sides by 2:
\( 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx = 0 \)
Group the terms strategically:
\( (x^2 + y^2 - 2xy) + (y^2 + z^2 - 2yz) + (z^2 + x^2 - 2zx) = 0 \)
Recognize each grouped term as a perfect square:
\( (x - y)^2 + (y - z)^2 + (z - x)^2 = 0 \)
Since each squared term is non-negative and their sum equals zero, each term must individually equal zero:
\( x - y = 0, \quad y - z = 0, \quad z - x = 0 \)
Therefore:
\( x = y = z \)
In simple words: Rearrange the condition, multiply by 2, regroup into perfect squares, and use the fact that a sum of squares can only be zero if each square is zero.

Exam Tip: This proof technique - converting to a sum of squares and using the non-negativity property - is common in algebra; watch for opportunities to group terms into \( (a - b)^2 \) form.

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