Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 4 Expansions

Exercise 4.1

Question 1. Expand each of the following expressions:
(i) \( (a + 4)(a + 7) \)
(ii) \( (m + 8)(m - 7) \)
(iii) \( (x - 5)(x - 4) \)
(iv) \( (3x + 4)(2x - 1) \)
(v) \( (2x - 5)(2x + 5)(2x - 3) \)
Answer:
(i) We will use the standard expansion identity \( (x + a)(x + b) = x^2 + (a + b)x + ab \). Here, \( x = a \), \( a = 4 \), and \( b = 7 \):
\( (a + 4)(a + 7) = a^2 + 4a + 7a + 28 = a^2 + 11a + 28 \)

(ii) Applying the expansion formula \( (x + a)(x - b) = x^2 + (a - b)x - ab \) where \( x = m \), \( a = 8 \), and \( b = 7 \):
\( (m + 8)(m - 7) = m^2 + 8m - 7m - 56 = m^2 + m - 56 \)

(iii) Using the formula for subtraction \( (x - a)(x - b) = x^2 - (a + b)x + ab \) where \( a = 5 \) and \( b = 4 \):
\( (x - 5)(x - 4) = x^2 - 5x - 4x + 20 = x^2 - 9x + 20 \)

(iv) By multiplying the binomial factors directly:
\( (3x + 4)(2x - 1) = 6x^2 - 3x + 8x - 4 = 6x^2 + 5x - 4 \)
Here, the formula \( (x + a)(x - b) = x^2 + (a - b)x - ab \) helps us compute the middle term.

(v) First, simplify the product of the first two terms using the difference of squares: \( (2x - 5)(2x + 5) = 4x^2 - 25 \). Next, multiply by the third factor:
\( (4x^2 - 25)(2x - 3) = 8x^3 - 12x^2 - 50x + 75 \)
In simple words: To expand these expressions, multiply each term in the first bracket by every term in the second bracket, then combine the terms that are alike.

Exam Tip: Always pay close attention to positive and negative signs when distributing multiplication over subtraction to avoid calculation errors.

Question 2. Expand the following squared expressions:
(a) \( (a + 3b)^2 \)
(b) \( (2p - 3q)^2 \)
(c) \( \left(2a + \frac{1}{2a}\right)^2 \)
(d) \( (x - 3y - 2z)^2 \)
Answer:
(a) By utilizing the binomial square identity \( (x + y)^2 = x^2 + 2xy + y^2 \), we obtain:
\( (a + 3b)^2 = a^2 + 2(a)(3b) + (3b)^2 = a^2 + 6ab + 9b^2 \)

(b) We use the binomial difference identity \( (x - y)^2 = x^2 - 2xy + y^2 \) to get:
\( (2p - 3q)^2 = (2p)^2 - 2(2p)(3q) + (3q)^2 = 4p^2 - 12pq + 9q^2 \)

(c) Applying the same squared identity for addition:
\( \left(2a + \frac{1}{2a}\right)^2 = (2a)^2 + 2(2a)\left(\frac{1}{2a}\right) + \left(\frac{1}{2a}\right)^2 = 4a^2 + 2 + \frac{1}{4a^2} \)

(d) Utilizing the square of a trinomial identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \):
\( (x - 3y - 2z)^2 = x^2 + (-3y)^2 + (-2z)^2 + 2(x)(-3y) + 2(-3y)(-2z) + 2(x)(-2z) \)

\( \implies (x - 3y - 2z)^2 = x^2 + 9y^2 + 4z^2 - 6xy + 12yz - 4xz \)
In simple words: To square a term like (A + B), use the formula A squared, plus B squared, plus two times A times B. For three terms inside the brackets, square each term first, then add two times every possible pair multiplied together.

Exam Tip: When squaring a negative term like \( (-3y)^2 \), remember that the result is always positive: \( 9y^2 \).

Question 3. Find the square of each of the following expressions:
(a) \( 9m - 2n \)
(b) \( 3p - 4q \)
(c) \( \frac{7x}{9y} - \frac{9y}{7x} \)
(d) \( 2a + 3b - 4c \)
Answer:
(a) Applying the identity for the square of a difference, \( (x - y)^2 = x^2 - 2xy + y^2 \):
\( (9m - 2n)^2 = (9m)^2 - 2(9m)(2n) + (-2n)^2 = 81m^2 - 36mn + 4n^2 \)

(b) Using the formula \( (x - y)^2 = x^2 - 2xy + y^2 \):
\( (3p - 4q)^2 = (3p)^2 - 2(3p)(4q) + (4q)^2 = 9p^2 - 24pq + 16q^2 \)

(c) We square the fractional terms using the binomial square of difference formula:
\( \left(\frac{7x}{9y} - \frac{9y}{7x}\right)^2 = \left(\frac{7x}{9y}\right)^2 - 2\left(\frac{7x}{9y}\right)\left(\frac{9y}{7x}\right) + \left(\frac{9y}{7x}\right)^2 = \frac{49x^2}{81y^2} - 2 + \frac{81y^2}{49x^2} \)

(d) Using the trinomial identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \):
\( (2a + 3b - 4c)^2 = (2a)^2 + (3b)^2 + (-4c)^2 + 2(2a)(3b) + 2(3b)(-4c) + 2(2a)(-4c) \)

\( \implies (2a + 3b - 4c)^2 = 4a^2 + 9b^2 + 16c^2 + 12ab - 24bc - 16ac \)
In simple words: When squaring subtraction terms, make sure the middle cross-product term has a negative sign. For fractions, the numerator and denominator square separately, and their cross-products cancel out.

Exam Tip: Be careful with the cross-multiplication term in fractional squares where \( a \cdot \frac{1}{a} = 1 \); only the coefficient 2 remains.

Question 4. Find the product of the following expressions using algebraic identities:
(i) \( (5x - 9)(5x + 9) \)
(ii) \( (2x + 3y)(2x - 3y) \)
(iii) \( (a + b - c)(a - b + c) \)
(iv) \( (x + y - 3)(x + y + 3) \)
(v) \( (1 + a)(1 - a)(1 + a^2) \)
(vi) \( \left(a + \frac{2}{a} - 1\right)\left(a - \frac{2}{a} - 1\right) \)
Answer:
(i) We apply the difference of squares identity, \( (a - b)(a + b) = a^2 - b^2 \):
\( (5x - 9)(5x + 9) = (5x)^2 - (9)^2 = 25x^2 - 81 \)

(ii) Applying the same difference of squares identity:
\( (2x + 3y)(2x - 3y) = (2x)^2 - (3y)^2 = 4x^2 - 9y^2 \)

(iii) We regroup the terms to fit the difference of squares format:
\( (a + b - c)(a - b + c) = [a + (b - c)][a - (b - c)] \)

\( \implies (a)^2 - (b - c)^2 \)
Expanding the second bracket:
\( \implies a^2 - (b^2 + c^2 - 2bc) = a^2 - b^2 - c^2 + 2bc \)

(iv) Grouping \( (x + y) \) together as a single term:
\( (x + y - 3)(x + y + 3) = (x + y)^2 - (3)^2 \)

\( \implies x^2 + y^2 + 2xy - 9 \)

(v) Multiply the first two brackets using the difference of squares, then multiply the result with the third bracket:
\( (1 + a)(1 - a)(1 + a^2) = [(1)^2 - (a)^2](1 + a^2) = (1 - a^2)(1 + a^2) \)

\( \implies (1)^2 - (a^2)^2 = 1 - a^4 \)

(vi) Grouping the terms \( (a - 1) \) together:
\( \left(a + \frac{2}{a} - 1\right)\left(a - \frac{2}{a} - 1\right) = \left[(a - 1) + \frac{2}{a}\right]\left[(a - 1) - \frac{2}{a}\right] \)

\( \implies (a - 1)^2 - \left(\frac{2}{a}\right)^2 = a^2 - 2a + 1 - \frac{4}{a^2} \)
In simple words: Look for pairs of terms that are added in one bracket and subtracted in the other. This lets you use the short-cut formula: square of first term minus square of second term.

Exam Tip: Grouping terms strategically (like combining \( a - 1 \) or \( x + y \)) allows you to apply the standard difference of squares formula very easily.

Question 5. Evaluate the squares of the following numbers using algebraic identities:
(a) \( (95)^2 \)
(b) \( (103)^2 \)
(c) \( (999)^2 \)
(d) \( (1005)^2 \)
Answer:
(a) We can write 95 as \( (100 - 5) \) and apply the formula \( (x - y)^2 = x^2 - 2xy + y^2 \):
\( (95)^2 = (100 - 5)^2 = (100)^2 - 2(100)(5) + (5)^2 \)

\( \implies (95)^2 = 10000 - 1000 + 25 = 9025 \)

(b) Writing 103 as \( (100 + 3) \) and using \( (x + y)^2 = x^2 + 2xy + y^2 \):
\( (103)^2 = (100 + 3)^2 = (100)^2 + 2(100)(3) + (3)^2 \)

\( \implies (103)^2 = 10000 + 600 + 9 = 10609 \)

(c) Expressing 999 as \( (1000 - 1) \):
\( (999)^2 = (1000 - 1)^2 = (1000)^2 - 2(1000)(1) + (1)^2 \)

\( \implies (999)^2 = 1000000 - 2000 + 1 = 998001 \)

(d) Rewriting 1005 as \( (1000 + 5) \):
\( (1005)^2 = (1000 + 5)^2 = (1000)^2 + 2(1000)(5) + (5)^2 \)

\( \implies (1005)^2 = 1000000 + 10000 + 25 = 1010025 \)
In simple words: Rewrite numbers close to 100 or 1000 as a sum or difference, then use basic algebraic formulas to square them easily without long multiplication.

Exam Tip: Rewriting numbers relative to base 100 or 1000 makes mental arithmetic much faster and prevents multiplication slip-ups.

Question 6. Find the value of the following products using identities:
(a) \( 399 \times 401 \)
(b) \( 999 \times 1001 \)
(c) \( 4.9 \times 5.1 \)
(d) \( 15.9 \times 16.1 \)
Answer:
(a) We can express the numbers as \( (400 - 1) \) and \( (400 + 1) \). Utilizing the identity \( (x - y)(x + y) = x^2 - y^2 \):
\( 399 \times 401 = (400 - 1)(400 + 1) = (400)^2 - (1)^2 \)

\( \implies 399 \times 401 = 160000 - 1 = 159999 \)

(b) Rewriting the values as \( (1000 - 1) \) and \( (1000 + 1) \):
\( 999 \times 1001 = (1000 - 1)(1000 + 1) = (1000)^2 - (1)^2 \)

\( \implies 999 \times 1001 = 1000000 - 1 = 999999 \)

(c) Writing the decimals as \( (5 - 0.1) \) and \( (5 + 0.1) \):
\( 4.9 \times 5.1 = (5 - 0.1)(5 + 0.1) = (5)^2 - (0.1)^2 \)

\( \implies 4.9 \times 5.1 = 25 - 0.01 = 24.99 \)

(d) Representing these decimals as \( (16 - 0.1) \) and \( (16 + 0.1) \):
\( 15.9 \times 16.1 = (16 - 0.1)(16 + 0.1) = (16)^2 - (0.1)^2 \)

\( \implies 15.9 \times 16.1 = 256 - 0.01 = 255.99 \)
In simple words: When multiplying two numbers that are equally spaced around a round number, rewrite them as (A - B) and (A + B) and then calculate A squared minus B squared.

Exam Tip: For decimal multiplications, ensure the decimal place is squared correctly; for example, \( (0.1)^2 = 0.01 \), not \( 0.1 \).

Question 7. If \( a - b = 10 \) and \( ab = 11 \), find the value of \( a + b \).
Answer:
We start by using the standard algebraic relationship:
\( (a - b)^2 = a^2 + b^2 - 2ab \)
Substituting the known values \( a - b = 10 \) and \( ab = 11 \):
\( (10)^2 = a^2 + b^2 - 2(11) \)

\( \implies 100 = a^2 + b^2 - 22 \)

\( \implies a^2 + b^2 = 100 + 22 = 122 \)

Next, we use the identity for the square of a sum:
\( (a + b)^2 = a^2 + b^2 + 2ab \)
Plugging in \( a^2 + b^2 = 122 \) and \( ab = 11 \):
\( (a + b)^2 = 122 + 2(11) = 122 + 22 = 144 \)
Taking the square root on both sides:

\( \implies a + b = \sqrt{144} = \pm 12 \)
In simple words: First, find the sum of squares of the numbers using the difference formula. Then use that result in the sum formula to get the final answer.

Exam Tip: Don't forget to write both the positive and negative signs (\( \pm \)) when taking the square root of a number.

Question 8. If \( x + y = 9 \) and \( xy = 20 \), find the value of:
(i) \( x^2 + y^2 \) and \( x - y \)
(ii) \( x^2 - y^2 \)
Answer:
(i) By squaring the sum, we know that:
\( (x + y)^2 = x^2 + y^2 + 2xy \)
Substituting the given values:
\( (9)^2 = x^2 + y^2 + 2(20) \)

\( \implies 81 = x^2 + y^2 + 40 \)

\( \implies x^2 + y^2 = 81 - 40 = 41 \)

Next, to find \( x - y \), we use the difference identity:
\( (x - y)^2 = x^2 + y^2 - 2xy \)
Substituting our known values:
\( (x - y)^2 = 41 - 2(20) = 41 - 40 = 1 \)

\( \implies x - y = \pm 1 \)

(ii) Utilizing the difference of squares identity:
\( x^2 - y^2 = (x - y)(x + y) \)
Plugging in the values we solved:

\( \implies x^2 - y^2 = (\pm 1)(9) = \pm 9 \)
In simple words: Find the sum of squares first, then use it to find the subtraction of the terms. Finally, multiply the sum and subtraction to get the difference of squares.

Exam Tip: Be mindful of the \( \pm \) in the difference \( x - y = \pm 1 \), as it directly affects the sign of the product \( x^2 - y^2 \).

Question 9. If \( a + \frac{1}{a} = 6 \), find the value of:
(i) \( a^2 + \frac{1}{a^2} \) and \( a - \frac{1}{a} \)
(ii) \( a^2 - \frac{1}{a^2} \)
Answer:
(i) We start by squaring the given equation:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2(a)\left(\frac{1}{a}\right) \)

\( \implies (6)^2 = a^2 + \frac{1}{a^2} + 2 \)

\( \implies 36 = a^2 + \frac{1}{a^2} + 2 \)

\( \implies a^2 + \frac{1}{a^2} = 34 \)

Now, we calculate the square of the difference term:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substituting \( a^2 + \frac{1}{a^2} = 34 \):
\( \left(a - \frac{1}{a}\right)^2 = 34 - 2 = 32 \)
Taking square root on both sides:

\( \implies a - \frac{1}{a} = \pm \sqrt{32} = \pm 4\sqrt{2} \)

(ii) Using the factorization formula for the difference of two squares:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)
Substituting the values we have:

\( \implies a^2 - \frac{1}{a^2} = (6)(\pm 4\sqrt{2}) = \pm 24\sqrt{2} \)
In simple words: First, square the sum formula to find the value of the squared terms. Use that value to find the subtraction formula, and then multiply the sum and difference to get the final difference of squares.

Exam Tip: Simplify square roots completely; express \( \sqrt{32} \) as \( 4\sqrt{2} \) to secure maximum marks.

Question 10. If \( a - \frac{1}{a} = 10 \), find the value of:
(i) \( a^2 + \frac{1}{a^2} \) and \( a + \frac{1}{a} \)
(ii) \( a^2 - \frac{1}{a^2} \)
Answer:
(i) We start by squaring the given difference:
\( \left(a - \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} - 2 \)
Substituting our value \( a - \frac{1}{a} = 10 \):
\( (10)^2 = a^2 + \frac{1}{a^2} - 2 \)

\( \implies 100 = a^2 + \frac{1}{a^2} - 2 \)

\( \implies a^2 + \frac{1}{a^2} = 102 \)

Next, to obtain the sum, we use the sum of squares identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
Substitute \( a^2 + \frac{1}{a^2} = 102 \):
\( \left(a + \frac{1}{a}\right)^2 = 102 + 2 = 104 \)
Taking square root on both sides:

\( \implies a + \frac{1}{a} = \sqrt{104} = \pm \sqrt{4 \times 26} = \pm 2\sqrt{26} \)

(ii) We now use the identity for the difference of two squares:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)
Substitute the known values:

\( \implies a^2 - \frac{1}{a^2} = (\pm 2\sqrt{26})(10) = \pm 20\sqrt{26} \)
In simple words: Square the given equation to find the value of the squared sum. Then add 2 to that result and take the root to find the sum term. Finally, multiply them.

Exam Tip: Be precise when simplifying roots like \( \sqrt{104} \). Look for perfect square factors like 4, which simplifies to \( 2\sqrt{26} \).

Question 11. If \( x + \frac{1}{x} = 3 \), find the value of:
(i) \( x^2 + \frac{1}{x^2} \)
(ii) \( x^4 + \frac{1}{x^4} \)
Answer:
(i) We start by squaring both sides of the given equation:
\( \left(x + \frac{1}{x}\right)^2 = (3)^2 \)

\( \implies x^2 + \frac{1}{x^2} + 2(x)\left(\frac{1}{x}\right) = 9 \)

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

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

(ii) Now we square both sides of the result \( x^2 + \frac{1}{x^2} = 7 \):
\( \left(x^2 + \frac{1}{x^2}\right)^2 = (7)^2 \)

\( \implies x^4 + \frac{1}{x^4} + 2 = 49 \)

\( \implies x^4 + \frac{1}{x^4} = 49 - 2 = 47 \)
In simple words: Square both sides of the first equation to get the power 2 term. Then square the result again to get the power 4 term.

Exam Tip: Remember that squaring \( x^2 + \frac{1}{x^2} \) produces a constant term of \( 2 \) because \( x^2 \cdot \frac{1}{x^2} = 1 \).

Question 12. If \( p + q = 8 \) and \( p - q = 4 \), find the value of:
(i) \( pq \)
(ii) \( p^2 + q^2 \)
Answer:
(i) First, we write the square of both equations:
\( (p + q)^2 = (8)^2 \implies p^2 + q^2 + 2pq = 64 \) - (1)
\( (p - q)^2 = (4)^2 \implies p^2 + q^2 - 2pq = 16 \) - (2)
Subtracting equation (2) from equation (1):
\( (p^2 + q^2 + 2pq) - (p^2 + q^2 - 2pq) = 64 - 16 \)

\( \implies 4pq = 48 \)

\( \implies pq = 12 \)

(ii) We can find \( p^2 + q^2 \) by adding equation (1) and (2) or by substituting the value of \( pq \):
Substitute \( pq = 12 \) into equation (1):
\( p^2 + q^2 + 2(12) = 64 \)

\( \implies p^2 + q^2 + 24 = 64 \)

\( \implies p^2 + q^2 = 64 - 24 = 40 \)
In simple words: Square both equations, subtract them to find pq, and use pq to solve for the sum of squares.

Exam Tip: Alternatively, you can add equations (1) and (2) directly to find \( p^2+q^2 \): \( 2(p^2+q^2) = 80 \implies p^2+q^2 = 40 \).

Question 13. Given \( m - n = 0.9 \) and \( mn = 0.36 \), find the value of:
a. \( m + n \)
b. \( m^2 - n^2 \)
Answer:
a. First, we find \( m^2 + n^2 \) using the squaring identity of difference:
\( (m - n)^2 = m^2 - 2mn + n^2 \)
Substituting our given values:
\( (0.9)^2 = m^2 + n^2 - 2(0.36) \)

\( \implies 0.81 = m^2 + n^2 - 0.72 \)

\( \implies m^2 + n^2 = 0.81 + 0.72 = 1.53 \)

Next, we compute \( (m + n)^2 \) using the sum squared identity:
\( (m + n)^2 = m^2 + 2mn + n^2 \)
Substituting our calculated and given values:
\( (m + n)^2 = 1.53 + 2(0.36) = 1.53 + 0.72 = 2.25 \)
Taking the square root on both sides:

\( \implies m + n = \pm 1.5 \)

b. To find \( m^2 - n^2 \), we multiply our results:
\( m^2 - n^2 = (m + n)(m - n) \)

\( \implies m^2 - n^2 = (\pm 1.5)(0.9) = \pm 1.35 \)
In simple words: Solve for the sum of squares first, use that to calculate the positive and negative values of m + n, then multiply that by the given subtraction to find the difference of squares.

Exam Tip: Be accurate with decimal calculations. Keep track of the decimal positions, especially when squaring and multiplying.

Question 14. If \( x + y = 1 \) and \( xy = -12 \), find the value of:
(i) \( x - y \)
(ii) \( x^2 - y^2 \)
Answer:
(i) Squaring the sum equation:
\( (x + y)^2 = (1)^2 \)

\( \implies x^2 + y^2 + 2xy = 1 \)
Substituting \( xy = -12 \):

\( \implies x^2 + y^2 + 2(-12) = 1 \)

\( \implies x^2 + y^2 - 24 = 1 \)

\( \implies x^2 + y^2 = 25 \)

We can now use the squared difference identity:
\( (x - y)^2 = x^2 + y^2 - 2xy \)

\( \implies (x - y)^2 = 25 - 2(-12) = 25 + 24 = 49 \)
Taking square root on both sides:

\( \implies x - y = \pm 7 \)

(ii) Applying the difference of squares identity:
\( x^2 - y^2 = (x + y)(x - y) \)

\( \implies x^2 - y^2 = (1)(\pm 7) = \pm 7 \)
In simple words: First square the sum of the variables, plug in the given product, and calculate the square of the difference. Use that to solve for the final subtraction and difference of squares.

Exam Tip: Be extremely careful when subtracting negative terms: \( -2(-12) \) becomes \( +24 \), which is a very common place to make a sign error.

Question 15. If \( a^2 - 7a + 1 = 0 \), find the value of:
(i) \( a + \frac{1}{a} \)
(ii) \( a^2 + \frac{1}{a^2} \)
Answer:
(i) Divide the entire quadratic equation by \( a \):
\( \frac{a^2}{a} - \frac{7a}{a} + \frac{1}{a} = \frac{0}{a} \)

\( \implies a - 7 + \frac{1}{a} = 0 \)

\( \implies a + \frac{1}{a} = 7 \)

(ii) Squaring the result:
\( \left(a + \frac{1}{a}\right)^2 = (7)^2 \)

\( \implies a^2 + \frac{1}{a^2} + 2 = 49 \)

\( \implies a^2 + \frac{1}{a^2} = 49 - 2 = 47 \)
In simple words: Divide the whole quadratic equation by the variable first. This rearranges it into a simple fractional sum, which can then be squared to find the second part.

Exam Tip: Whenever you see a quadratic equation of the form \( ax^2 + bx + c = 0 \) and you need to find fractional powers, divide by \( x \) as your very first step.

Question 16. If \( a^2 - 3a - 1 = 0 \), find the value of:
(i) \( a - \frac{1}{a} \)
(ii) \( a + \frac{1}{a} \)
(iii) \( a^2 - \frac{1}{a^2} \)
Answer:
(i) Dividing the given expression by \( a \):
\( a - 3 - \frac{1}{a} = 0 \)

\( \implies a - \frac{1}{a} = 3 \)

(ii) Squaring both sides of this equation:
\( \left(a - \frac{1}{a}\right)^2 = (3)^2 \)

\( \implies a^2 + \frac{1}{a^2} - 2 = 9 \)

\( \implies a^2 + \frac{1}{a^2} = 11 \)

Next, using the squared sum identity:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 = 11 + 2 = 13 \)
Taking square root:

\( \implies a + \frac{1}{a} = \pm \sqrt{13} \)

(iii) Finally, we use the difference of squares identity:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)

\( \implies a^2 - \frac{1}{a^2} = (\pm \sqrt{13})(3) = \pm 3\sqrt{13} \)
In simple words: Divide the starting equation by the variable to get the difference. Use that to find the sum of squares, then find the sum, and finally multiply them to get the difference of squares.

Exam Tip: Be sure to keep the square root as an exact radical expression (\( \sqrt{13} \)) since it is an irrational number.

Question 17. If \( 2x + 3y = 10 \) and \( xy = 5 \), find the value of \( 4x^2 + 9y^2 \).
Answer:
We can rewrite \( 4x^2 + 9y^2 \) as \( (2x)^2 + (3y)^2 \). Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \):
\( 4x^2 + 9y^2 = (2x + 3y)^2 - 2(2x)(3y) \)

\( \implies 4x^2 + 9y^2 = (2x + 3y)^2 - 12xy \)
Substituting our known values \( 2x + 3y = 10 \) and \( xy = 5 \):

\( \implies 4x^2 + 9y^2 = (10)^2 - 12(5) = 100 - 60 = 40 \)
In simple words: Square the whole given linear expression, subtract the cross term multiplied by 12, and you will get the final answer.

Exam Tip: When subtracting the cross-product term, make sure to multiply by the coefficients of both variables: \( 2(2x)(3y) = 12xy \), not just \( 2xy \).

Question 18. If \( x + y + z = 12 \) and \( xy + yz + zx = 27 \), find the value of \( x^2 + y^2 + z^2 \).
Answer:
By squaring the trinomial sum equation:
\( (x + y + z)^2 = (12)^2 \)
Using the expansion formula \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \):
\( x^2 + y^2 + z^2 + 2(xy + yz + zx) = 144 \)
Substituting \( xy + yz + zx = 27 \):

\( \implies x^2 + y^2 + z^2 + 2(27) = 144 \)

\( \implies x^2 + y^2 + z^2 + 54 = 144 \)

\( \implies x^2 + y^2 + z^2 = 144 - 54 = 90 \)
In simple words: Square the sum of the three variables, subtract two times the sum of their pairwise products, and that gives you the sum of their individual squares.

Exam Tip: Factor the 2 out of the cross-terms: \( 2xy + 2yz + 2zx = 2(xy + yz + zx) \) to make substitution easier.

Question 19. If \( a + b + c = 9 \) and \( a^2 + b^2 + c^2 = 41 \), find the value of \( ab + bc + ca \).
Answer:
We square both sides of the sum equation:
\( (a + b + c)^2 = (9)^2 \)

\( \implies a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81 \)
Substituting \( a^2 + b^2 + c^2 = 41 \):

\( \implies 41 + 2(ab + bc + ca) = 81 \)

\( \implies 2(ab + bc + ca) = 81 - 41 = 40 \)

\( \implies ab + bc + ca = 20 \)
In simple words: Square the sum of the three numbers, subtract the sum of their squares, and then divide the remainder by 2 to get the pairwise sum.

Exam Tip: Always show intermediate steps of shifting and dividing by 2 clearly to ensure full marks for the working.

Question 20. If \( p^2 + q^2 + r^2 = 64 \) and \( pq + qr + rp = 18 \), find the value of \( p + q + r \).
Answer:
We use the standard square of a trinomial identity:
\( (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + qr + rp) \)
Substituting the given values:
\( (p + q + r)^2 = 64 + 2(18) \)

\( \implies (p + q + r)^2 = 64 + 36 = 100 \)
Taking square root on both sides:

\( \implies p + q + r = \sqrt{100} = \pm 10 \)
In simple words: Add the sum of squares and two times the product-sum, then find the square root of that total.

Exam Tip: Remember to express the square root result as both positive and negative values (\( \pm 10 \)).

Question 21. If \( x + y + z = p \) and \( xy + yz + zx = q \), find the value of \( x^2 + y^2 + z^2 \) in terms of \( p \) and \( q \).
Answer:
Using the expansion of a trinomial squared:
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \)
Substitute the variables \( p \) and \( q \) into this identity:
\( (p)^2 = x^2 + y^2 + z^2 + 2(q) \)
Rearranging to find the sum of squares:

\( \implies x^2 + y^2 + z^2 = p^2 - 2q \)
In simple words: Write out the basic trinomial identity and plug in the letters p and q in their proper places.

Exam Tip: This is a formula derivation question, so make sure to write down the final algebraic expression clearly.

Exercise 4.2

Question 1. Expand the following cubic expressions using standard identities:
(i) \( (2a - 5b)^3 \)
(ii) \( (4x + 7y)^3 \)
(iii) \( \left(3a + \frac{1}{3a}\right)^3 \)
(iv) \( \left(4p - \frac{1}{p}\right)^3 \)
(v) \( \left(\frac{2m}{3n} + \frac{3n}{2m}\right)^3 \)
(vi) \( \left(a - \frac{1}{a} + b\right)^3 \)
Answer:
(i) Applying the cubic subtraction identity \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \):
\( (2a - 5b)^3 = (2a)^3 - (5b)^3 - 3(2a)(5b)(2a - 5b) \)

\( \implies 8a^3 - 125b^3 - 30ab(2a - 5b) \)

\( \implies 8a^3 - 125b^3 - 60a^2b + 150ab^2 \)

(ii) Applying the cubic addition identity \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \):
\( (4x + 7y)^3 = (4x)^3 + (7y)^3 + 3(4x)(7y)(4x + 7y) \)

\( \implies 64x^3 + 343y^3 + 84xy(4x + 7y) \)

\( \implies 64x^3 + 343y^3 + 336x^2y + 588xy^2 \)

(iii) Expanding with the fractional base in addition:
\( \left(3a + \frac{1}{3a}\right)^3 = (3a)^3 + \left(\frac{1}{3a}\right)^3 + 3(3a)\left(\frac{1}{3a}\right)\left(3a + \frac{1}{3a}\right) \)

\( \implies 27a^3 + \frac{1}{27a^3} + 3\left(3a + \frac{1}{3a}\right) \)

\( \implies 27a^3 + \frac{1}{27a^3} + 9a + \frac{1}{a} \)

(iv) Applying the difference cube identity with fractions:
\( \left(4p - \frac{1}{p}\right)^3 = (4p)^3 - \left(\frac{1}{p}\right)^3 - 3(4p)\left(\frac{1}{p}\right)\left(4p - \frac{1}{p}\right) \)

\( \implies 64p^3 - \frac{1}{p^3} - 12\left(4p - \frac{1}{p}\right) \)

\( \implies 64p^3 - \frac{1}{p^3} - 48p + \frac{12}{p} \)

(v) Solving the fractional terms:
\( \left(\frac{2m}{3n} + \frac{3n}{2m}\right)^3 = \left(\frac{2m}{3n}\right)^3 + \left(\frac{3n}{2m}\right)^3 + 3\left(\frac{2m}{3n}\right)\left(\frac{3n}{2m}\right)\left(\frac{2m}{3n} + \frac{3n}{2m}\right) \)

\( \implies \frac{8m^3}{27n^3} + \frac{27n^3}{8m^3} + 3\left(\frac{2m}{3n} + \frac{3n}{2m}\right) \)

\( \implies \frac{8m^3}{27n^3} + \frac{27n^3}{8m^3} + \frac{2m}{n} + \frac{9n}{2m} \)

(vi) By using the trinomial cubic expansion identity \( (a+b+c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2a + 3b^2c + 3c^2a + 3c^2b + 6abc \):
\( \left(a - \frac{1}{a} + b\right)^3 = a^3 + \left(-\frac{1}{a}\right)^3 + b^3 + 3a^2\left(-\frac{1}{a}\right) + 3a^2(b) + 3\left(-\frac{1}{a}\right)^2(a) + 3\left(-\frac{1}{a}\right)^2(b) + 3b^2(a) + 3b^2\left(-\frac{1}{a}\right) + 6(a)\left(-\frac{1}{a}\right)(b) \)

\( \implies a^3 - \frac{1}{a^3} + b^3 - 3a + 3a^2b + \frac{3}{a} + \frac{3b}{a^2} + 3ab^2 - \frac{3b^2}{a} - 6b \)
In simple words: To expand a bracket to the power of three, cube the first and second terms, then add or subtract three times the product of those terms multiplied by the entire original bracket.

Exam Tip: Keep your expansion signs steady. When cubing negative terms, the negative sign persists, but when squaring negative terms, they become positive.

Question 2. If \( 5x + \frac{1}{5x} = 7 \), find the value of \( 125x^3 + \frac{1}{125x^3} \).
Answer:
By utilizing the cubic identity for reciprocal terms: \( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Letting \( a = 5x \), we have:
\( \left(5x + \frac{1}{5x}\right)^3 = (5x)^3 + \left(\frac{1}{5x}\right)^3 + 3\left(5x + \frac{1}{5x}\right) \)
Substituting our given value of \( 7 \):
\( (7)^3 = 125x^3 + \frac{1}{125x^3} + 3(7) \)

\( \implies 343 = 125x^3 + \frac{1}{125x^3} + 21 \)

\( \implies 125x^3 + \frac{1}{125x^3} = 343 - 21 = 322 \)
In simple words: Cube the given equation. The cross-multiplied middle term simplifies down to a constant, making it easy to subtract and solve.

Exam Tip: Recognize that \( (5x)^3 \) evaluates to \( 125x^3 \); do not write \( 5x^3 \).

Question 3. If \( 3x - \frac{1}{3x} = 9 \), find the value of \( 27x^3 - \frac{1}{27x^3} \).
Answer:
By utilizing the cubic identity for reciprocal subtraction: \( \left(a - \frac{1}{a}\right)^3 = a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right) \)
Setting \( a = 3x \), we obtain:
\( \left(3x - \frac{1}{3x}\right)^3 = (3x)^3 - \left(\frac{1}{3x}\right)^3 - 3\left(3x - \frac{1}{3x}\right) \)
Substituting the given value of \( 9 \):
\( (9)^3 = 27x^3 - \frac{1}{27x^3} - 3(9) \)

\( \implies 729 = 27x^3 - \frac{1}{27x^3} - 27 \)

\( \implies 27x^3 - \frac{1}{27x^3} = 729 + 27 = 756 \)
In simple words: Cube the given equation. For a subtraction cube, you will add the constant term to the other side to get the final answer.

Exam Tip: Ensure you square or cube the coefficients as well: \( (3x)^3 = 27x^3 \), not \( 3x^3 \).

Question 4. If \( x + \frac{1}{x} = 5 \), find the value of:
(i) \( x^2 + \frac{1}{x^2} \)
(ii) \( x^3 + \frac{1}{x^3} \)
(iii) \( x^4 + \frac{1}{x^4} \)
Answer:
(i) Squaring both sides of our starting equation:
\( \left(x + \frac{1}{x}\right)^2 = (5)^2 \)

\( \implies x^2 + \frac{1}{x^2} + 2 = 25 \)

\( \implies x^2 + \frac{1}{x^2} = 23 \) - (2)

(ii) Now, we cube both sides of the starting equation:
\( \left(x + \frac{1}{x}\right)^3 = (5)^3 \)

\( \implies x^3 + \frac{1}{x^3} + 3\left(x + \frac{1}{x}\right) = 125 \)
Substituting \( x + \frac{1}{x} = 5 \):

\( \implies x^3 + \frac{1}{x^3} + 3(5) = 125 \)

\( \implies x^3 + \frac{1}{x^3} = 125 - 15 = 110 \)

(iii) Squaring both sides of equation (2):
\( \left(x^2 + \frac{1}{x^2}\right)^2 = (23)^2 \)

\( \implies x^4 + \frac{1}{x^4} + 2 = 529 \)

\( \implies x^4 + \frac{1}{x^4} = 529 - 2 = 527 \)
In simple words: Square the starting equation to get the power 2 term. Cube the starting equation to find the power 3 term. Finally, square the power 2 result to find the power 4 term.

Exam Tip: Be sure not to mix up the steps of cubing and squaring. Write down the formulas clearly before doing arithmetic.

Question 5. If \( a - \frac{1}{a} = 7 \), find the value of \( a^2 - \frac{1}{a^2} \).
Answer:
First, we square both sides of the given equation:
\( \left(a - \frac{1}{a}\right)^2 = (7)^2 \)

\( \implies a^2 + \frac{1}{a^2} - 2 = 49 \)

\( \implies a^2 + \frac{1}{a^2} = 51 \)

Next, we compute the square of the sum term:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
Substitute our calculated sum of squares:
\( \left(a + \frac{1}{a}\right)^2 = 51 + 2 = 53 \)
Taking the square root on both sides:

\( \implies a + \frac{1}{a} = \pm \sqrt{53} \)

Now, we utilize the difference of squares identity:
\( a^2 - \frac{1}{a^2} = \left(a + \frac{1}{a}\right)\left(a - \frac{1}{a}\right) \)
Substitute the known factors:

\( \implies a^2 - \frac{1}{a^2} = (\pm \sqrt{53})(7) = \pm 7\sqrt{53} \)
In simple words: First square the given difference to find the value of the squared terms. Add 2 to that result to find the square of the sum. Root it, and then multiply the sum and difference terms.

Exam Tip: Since 53 is a prime number, leave \( \sqrt{53} \) as a surd in the final answer.

Question 6. If \( a^2 + \frac{1}{a^2} = 14 \), find the value of \( a + \frac{1}{a} \).
Answer:
Using the identity for squaring a sum of reciprocal terms:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2} \)
Rearranging the terms on the right side:
\( \left(a + \frac{1}{a}\right)^2 = \left(a^2 + \frac{1}{a^2}\right) + 2 \)
Substituting the given value of \( 14 \):
\( \left(a + \frac{1}{a}\right)^2 = 14 + 2 = 16 \)
Taking square root on both sides:

\( \implies a + \frac{1}{a} = \pm 4 \)
In simple words: Add 2 to the given value of the squares, then find the square root of that sum to get the final answer.

Exam Tip: Always include both positive and negative values (\( \pm 4 \)) when taking the square root of a number, unless a specific domain restriction is given.

Question 6B. If \( a^2 + \frac{1}{a^2} = 14 \), find the value of \( a^3 + \frac{1}{a^3} \).
Answer: We can begin by using the algebraic identity \( (a+b)^2 = a^2 + 2ab + b^2 \).
By applying this, we have:
\( (a + \frac{1}{a})^2 = a^2 + 2(a)\left(\frac{1}{a}\right) + \left(\frac{1}{a}\right)^2 \)
\( \implies (a + \frac{1}{a})^2 = a^2 + 2 + \frac{1}{a^2} \)
\( \implies (a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2 \)
Substituting the given value \( a^2 + \frac{1}{a^2} = 14 \):
\( \implies (a + \frac{1}{a})^2 = 14 + 2 \)
\( \implies (a + \frac{1}{a})^2 = 16 \)
Taking the square root on both sides yields:
\( \implies a + \frac{1}{a} = \pm 4 \)
Next, we employ the identity \( a^3 + b^3 = (a+b)(a^2 + b^2 - ab) \):
\( a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)\left(a^2 + \frac{1}{a^2} - 1\right) \)
Substituting the known values into the equation:
\( = (\pm 4)(14 - 1) \)
\( = (\pm 4)(13) \)
\( = \pm 52 \)
In simple words: First, find the value of \( a + \frac{1}{a} \) by squaring it and substituting the given value. Then, use the sum of cubes formula to calculate the final answer.

Exam Tip: Remember to include both the positive and negative square roots (\( \pm 4 \)) so that you do not miss either of the two possible final answers.

Question 7. If \( m^2 + \frac{1}{m^2} = 51 \), find the value of \( m^3 - \frac{1}{m^3} \).
Answer: We are given that \( m^2 + \frac{1}{m^2} = 51 \).
Recall the algebraic expansion:
\( \left(m - \frac{1}{m}\right)^2 = m^2 + \frac{1}{m^2} - 2 \)
Substituting the given value of \( m^2 + \frac{1}{m^2} \):
\( \implies \left(m - \frac{1}{m}\right)^2 = 51 - 2 \)
\( \implies \left(m - \frac{1}{m}\right)^2 = 49 \)
\( \implies m - \frac{1}{m} = 7 \)
Now, we cube both sides of the relation:
\( \implies \left(m - \frac{1}{m}\right)^3 = 7^3 \)
Using the identity \( (a-b)^3 = a^3 - b^3 - 3ab(a-b) \), we expand the left side:
\( \implies m^3 - \frac{1}{m^3} - 3\left(m - \frac{1}{m}\right) = 343 \)
Substituting \( m - \frac{1}{m} = 7 \):
\( \implies m^3 - \frac{1}{m^3} - 3 \times 7 = 343 \)
\( \implies m^3 - \frac{1}{m^3} = 343 + 21 \)
\( \implies m^3 - \frac{1}{m^3} = 364 \)
In simple words: First, find \( m - \frac{1}{m} \) by subtracting 2 from the given equation and taking the square root. After that, cube both sides and substitute the values to find the required result.

Exam Tip: Be careful with the signs when using the cubed difference identity; the term \( -3(m - \frac{1}{m}) \) becomes positive when moved to the other side of the equation.

Question 8. If \( 9a^2 + \frac{1}{9a^2} = 23 \), find the value of \( 27a^3 + \frac{1}{27a^3} \).
Answer: We start with the given equation \( 9a^2 + \frac{1}{9a^2} = 23 \).
Consider the square of \( 3a + \frac{1}{3a} \):
\( \left(3a + \frac{1}{3a}\right)^2 = (3a)^2 + \left(\frac{1}{3a}\right)^2 + 2(3a)\left(\frac{1}{3a}\right) \)
\( \implies \left(3a + \frac{1}{3a}\right)^2 = 9a^2 + \frac{1}{9a^2} + 2 \)
Putting \( 9a^2 + \frac{1}{9a^2} = 23 \) into the relation:
\( \left(3a + \frac{1}{3a}\right)^2 = 23 + 2 = 25 \)
Taking the positive square root gives:
\( \implies 3a + \frac{1}{3a} = 5 \)
Now, we perform cubing on both sides of this equation:
\( \left(3a\right)^3 + \left(\frac{1}{3a}\right)^3 + 3(3a)\left(\frac{1}{3a}\right)\left(3a + \frac{1}{3a}\right) = (5)^3 \)
\( \implies 27a^3 + \frac{1}{27a^3} + 3(5) = 125 \)
\( \implies 27a^3 + \frac{1}{27a^3} = 125 - 15 \)
\( \implies 27a^3 + \frac{1}{27a^3} = 110 \)
In simple words: Express the given term as a perfect square to find the value of \( 3a + \frac{1}{3a} \). Then, raise that result to the power of three to isolate the required sum of cubes.

Exam Tip: Always verify that the cross-multiplication term \( 3 \times 3a \times \frac{1}{3a} \) simplifies to a constant, which in this case is simply 3.

Question 9. If \( x^2 + \frac{1}{x^2} = 18 \), find the value of:
(i) \( x - \frac{1}{x} \)
(ii) \( x^3 - \frac{1}{x^3} \)
Answer:
(i) We start by using the standard algebraic relationship:
\( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
Substituting \( x^2 + \frac{1}{x^2} = 18 \):
\( \implies \left(x - \frac{1}{x}\right)^2 = 18 - 2 = 16 \)
Taking the positive square root on both sides:
\( \implies x - \frac{1}{x} = 4 \)
(ii) To find the cubed difference, we expand \( \left(x - \frac{1}{x}\right)^3 \):
\( \left(x - \frac{1}{x}\right)^3 = x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right) \)
Substituting the values we know, \( x - \frac{1}{x} = 4 \) and \( 4^3 = 64 \):
\( \implies 64 = x^3 - \frac{1}{x^3} - 3(4) \)
\( \implies x^3 - \frac{1}{x^3} = 64 + 12 \)
\( \implies x^3 - \frac{1}{x^3} = 76 \)
In simple words: First, subtract 2 from the given sum of squares to find \( x - \frac{1}{x} \). Then, use the cubed difference formula and insert the first result to solve the second part.

Exam Tip: Keep your steps separate and clearly label parts (i) and (ii) to ensure the examiner can easily follow your derivation.

Question 10. If \( p + \frac{1}{p} = 6 \), find the value of:
(i) \( p^2 + \frac{1}{p^2} \)
(ii) \( p^4 + \frac{1}{p^4} \)
(iii) \( p^3 + \frac{1}{p^3} \)
Answer:
(i) Let us square the given equation \( p + \frac{1}{p} = 6 \):
\( \left(p + \frac{1}{p}\right)^2 = p^2 + \frac{1}{p^2} + 2 \)
\( \implies 36 = p^2 + \frac{1}{p^2} + 2 \)
\( \implies p^2 + \frac{1}{p^2} = 36 - 2 = 34 \)
(ii) Next, we square the result obtained in part (i):
\( \left(p^2 + \frac{1}{p^2}\right)^2 = p^4 + \frac{1}{p^4} + 2 \)
\( \implies (34)^2 = p^4 + \frac{1}{p^4} + 2 \)
\( \implies 1156 = p^4 + \frac{1}{p^4} + 2 \)
\( \implies p^4 + \frac{1}{p^4} = 1156 - 2 = 1154 \)
(iii) Finally, to get the sum of cubes, we expand \( \left(p + \frac{1}{p}\right)^3 \):
\( \left(p + \frac{1}{p}\right)^3 = p^3 + \frac{1}{p^3} + 3\left(p + \frac{1}{p}\right) \)
\( \implies 216 = p^3 + \frac{1}{p^3} + 3(6) \)
\( \implies p^3 + \frac{1}{p^3} = 216 - 18 = 198 \)
In simple words: Square the initial equation to solve part (i). Then, square that answer to find the power of four in part (ii). For part (iii), cube the very first equation.

Exam Tip: Be sure to compute \( 34^2 \) accurately as 1156 to avoid minor calculation errors in the middle of your work.

Question 11. If \( r - \frac{1}{r} = 4 \), find the value of:
(i) \( r^2 + \frac{1}{r^2} \)
(ii) \( r^4 + \frac{1}{r^4} \)
(iii) \( r^3 - \frac{1}{r^3} \)
Answer:
(i) Let us square both sides of \( r - \frac{1}{r} = 4 \):
\( \left(r - \frac{1}{r}\right)^2 = r^2 + \frac{1}{r^2} - 2 \)
\( \implies (4)^2 = r^2 + \frac{1}{r^2} - 2 \)
\( \implies r^2 + \frac{1}{r^2} = 16 + 2 = 18 \)
(ii) Now, we square the result from part (i):
\( \left(r^2 + \frac{1}{r^2}\right)^2 = r^4 + \frac{1}{r^4} + 2 \)
\( \implies (18)^2 = r^4 + \frac{1}{r^4} + 2 \)
\( \implies r^4 + \frac{1}{r^4} = 324 - 2 = 322 \)
(iii) To find the sum/difference of cubes, we expand \( \left(r - \frac{1}{r}\right)^3 \):
\( \left(r - \frac{1}{r}\right)^3 = r^3 - \frac{1}{r^3} - 3\left(r - \frac{1}{r}\right) \)
\( \implies (4)^3 = r^3 - \frac{1}{r^3} - 3(4) \)
\( \implies r^3 - \frac{1}{r^3} = 64 + 12 = 76 \)
In simple words: Square the given equation to find the sum of squares. Square that new result to get the fourth-power sum. Finally, use the cubed difference identity with the initial value to solve the third part.

Exam Tip: Remember that squaring a negative term like \( -2 \) inside the square of a difference yields a positive term when rearranged on the other side.

Question 12. If \( a + \frac{1}{a} = 2 \), show that \( a^2 + \frac{1}{a^2} = a^3 + \frac{1}{a^3} = a^4 + \frac{1}{a^4} \).
Answer: We are given that \( a + \frac{1}{a} = 2 \).
First, squaring both sides of this expression:
\( \left(a + \frac{1}{a}\right)^2 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies (2)^2 = a^2 + \frac{1}{a^2} + 2 \)
\( \implies a^2 + \frac{1}{a^2} = 4 - 2 = 2 \)
Second, let us cube both sides of the original expression:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
\( \implies (2)^3 = a^3 + \frac{1}{a^3} + 3(2) \)
\( \implies a^3 + \frac{1}{a^3} = 8 - 6 = 2 \)
Third, we square the equation obtained for \( a^2 + \frac{1}{a^2} = 2 \):
\( \left(a^2 + \frac{1}{a^2}\right)^2 = a^4 + \frac{1}{a^4} + 2 \)
\( \implies (2)^2 = a^4 + \frac{1}{a^4} + 2 \)
\( \implies a^4 + \frac{1}{a^4} = 4 - 2 = 2 \)
Comparing these three final values, we observe:
\( a^2 + \frac{1}{a^2} = a^3 + \frac{1}{a^3} = a^4 + \frac{1}{a^4} = 2 \)
In simple words: By separately squaring, cubing, and squaring again, we show that all three algebraic expressions equal exactly 2.

Exam Tip: Clearly state at the end that all three expressions evaluate to 2, which completes the required proof.

Question 13. If \( x + \frac{1}{x} = p \) and \( x - \frac{1}{x} = q \), show that \( p^2 - q^2 = 4 \).
Answer: We are given the two equations \( x + \frac{1}{x} = p \) and \( x - \frac{1}{x} = q \).
Squaring the first equation:
\( \left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies p^2 = x^2 + \frac{1}{x^2} + 2 \)
\( \implies x^2 + \frac{1}{x^2} = p^2 - 2 \) ..... (1)
Now, squaring the second equation:
\( \left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies q^2 = x^2 + \frac{1}{x^2} - 2 \)
\( \implies x^2 + \frac{1}{x^2} = q^2 + 2 \) ..... (2)
Since equations (1) and (2) both represent the value of \( x^2 + \frac{1}{x^2} \), we can set them equal to each other:
\( p^2 - 2 = q^2 + 2 \)
Rearranging the terms to isolate the variables:
\( \implies p^2 - q^2 = 4 \)
In simple words: Square both equations to find two different expressions for \( x^2 + \frac{1}{x^2} \). Setting those two equal to each other gives the required relation.

Exam Tip: Ensure that you number your equations (1) and (2) so you can reference them clearly when setting them equal.

Question 14. If \( a + \frac{1}{a} = p \), find the value of \( a^3 + \frac{1}{a^3} \).
Answer: Given that \( a + \frac{1}{a} = p \).
Let us cube both sides of this equation:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Substituting \( a + \frac{1}{a} = p \) into the expanded formula:
\( \implies p^3 = a^3 + \frac{1}{a^3} + 3(p) \)
Isolating the cubed terms on one side:
\( \implies a^3 + \frac{1}{a^3} = p^3 - 3p \)
Factoring out \( p \):
\( = p(p^2 - 3) \)
In simple words: Cube the given equation and then subtract \( 3p \) from both sides. You can also pull out the common factor \( p \) at the end.

Exam Tip: Factoring the final expression as \( p(p^2 - 3) \) is a neat way to write the final simplified answer and shows solid algebraic skills.

Question 15. If \( \left(a + \frac{1}{a}\right)^2 = 3 \), find the value of \( a^3 + \frac{1}{a^3} \).
Answer: We are given that \( \left(a + \frac{1}{a}\right)^2 = 3 \).
Taking the positive square root on both sides:
\( \implies a + \frac{1}{a} = \sqrt{3} \)
Now, we write the identity for the sum of cubes:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Substituting \( a + \frac{1}{a} = \sqrt{3} \):
\( \implies (\sqrt{3})^3 = a^3 + \frac{1}{a^3} + 3(\sqrt{3}) \)
Since \( (\sqrt{3})^3 = 3\sqrt{3} \):
\( \implies a^3 + \frac{1}{a^3} = 3\sqrt{3} - 3\sqrt{3} = 0 \)
In simple words: Find the square root of the given equation to get \( \sqrt{3} \). Cube both sides, and you will see the terms subtract to zero.

Exam Tip: Be comfortable working with surds; remember that \( (\sqrt{3})^3 = \sqrt{3} \times \sqrt{3} \times \sqrt{3} = 3\sqrt{3} \).

Question 16. If \( a + b + c = 0 \), prove that \( a^3 + b^3 + c^3 = 3abc \).
Answer: We start with the given equation:
\( a + b + c = 0 \) ..... (i)
This can be written as:
\( \implies (a+b) + c = 0 \ )
Now, let us perform cubing on both sides:
\( \implies (a+b)^3 + c^3 + 3(a+b)(c)(a+b+c) = 0 \)
Since \( a+b+c = 0 \), the last term on the left side becomes 0:
\( \implies a^3 + b^3 + 3ab(a+b) + c^3 + 0 = 0 \)
\( \implies a^3 + b^3 + c^3 + 3ab(a+b) = 0 \) ..... (2)
From equation (i), we know that \( a+b = -c \). Substituting this into (2):
\( a^3 + b^3 + c^3 + 3ab(-c) = 0 \)
\( \implies a^3 + b^3 + c^3 - 3abc = 0 \)
\( \implies a^3 + b^3 + c^3 = 3abc \)
In simple words: Group two variables together and cube both sides. Use the original sum to simplify the cross term, which reveals the required formula.

Exam Tip: This is a very standard and important identity. Memorizing the derivation steps can help you tackle similar multi-variable algebraic proofs quickly.

Question 17. If \( a + 2b + c = 0 \), prove that \( a^3 + 8b^3 + c^3 = 6abc \).
Answer: We are given the relation:
\( a + 2b + c = 0 \) ..... (i)
This can be rewritten as:
\( \implies (a + 2b) + c = 0 \)
Cubing both sides of this equation:
\( \implies (a+2b)^3 + c^3 + 3(a+2b)c(a+2b+c) = 0 \)
Expanding \( (a+2b)^3 \) and substituting \( a+2b+c = 0 \) into the third term:
\( \implies a^3 + 8b^3 + 6ab(a+2b) + c^3 + 0 = 0 \)
\( \implies a^3 + 8b^3 + c^3 + 6ab(a+2b) = 0 \) ..... (2)
From equation (i), we can see that \( a + 2b = -c \). Let us substitute this into (2):
\( a^3 + 8b^3 + 6ab(-c) = 0 \)
\( \implies a^3 + 8b^3 + c^3 - 6abc = 0 \)
\( \implies a^3 + 8b^3 + c^3 = 6abc \)
In simple words: Group \( a + 2b \) together and cube the entire equation. Replace the sum with \( -c \) to complete the proof.

Exam Tip: Pay attention to coefficients during expansion: \( (2b)^3 = 8b^3 \) and \( 3 \times a \times 2b = 6ab \).

Question 18. If \( x^3 + y^3 = 9 \) and \( x + y = 3 \), find the value of \( xy \).
Answer: We are given \( x^3 + y^3 = 9 \) and \( x+y = 3 \).
Let us write down the cubed sum formula:
\( (x + y)^3 = x^3 + y^3 + 3xy(x+y) \)
Substitute the given values into this equation:
\( \implies (3)^3 = 9 + 3xy(3) \)
\( \implies 27 = 9 + 9xy \)
Subtracting 9 from both sides:
\( \implies 9xy = 27 - 9 = 18 \)
Dividing by 9:
\( \implies xy = 2 \)
In simple words: Put the given values directly into the standard cubed sum formula, then solve for the unknown product \( xy \).

Exam Tip: Be systematic in your arithmetic; make sure to perform subtraction before dividing to avoid errors.

Question 19. If \( a+b = 5 \) and \( ab = 2 \), find the value of \( a^3 + b^3 \).
Answer: We can find the sum of squares first using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \):
\( a^2 + b^2 = (a+b)^2 - 2ab \)
Substituting \( a+b = 5 \) and \( ab = 2 \):
\( \implies a^2 + b^2 = (5)^2 - 2(2) \)
\( \implies a^2 + b^2 = 25 - 4 \)
\( \implies a^2 + b^2 = 21 \)
Next, we use the sum of cubes identity \( a^3 + b^3 = (a+b)(a^2 + b^2 - ab) \):
\( a^3 + b^3 = (a+b)(a^2 + b^2 - ab) \)
Substituting the values we have calculated and those given:
\( = (5)(21 - 2) \)
\( = (5)(19) \)
\( = 95 \)
In simple words: Find \( a^2 + b^2 \) first by squaring the sum and subtracting twice the product. Then, use that in the sum of cubes formula.

Exam Tip: You can also solve this in one step using the formula \( a^3 + b^3 = (a+b)^3 - 3ab(a+b) \), which is often quicker.

Question 20. If \( p - q = -1 \) and \( pq = -12 \), find the value of \( p^3 - q^3 \).
Answer: We are given that \( p - q = -1 \) and \( pq = -12 \).
We use the cubed difference formula:
\( (p - q)^3 = p^3 - q^3 - 3pq(p - q) \)
Substituting the given values into the formula:
\( \implies (-1)^3 = p^3 - q^3 - 3(-12)(-1) \)
\( \implies -1 = p^3 - q^3 - 36 \)
Rearranging the terms to find \( p^3 - q^3 \):
\( \implies p^3 - q^3 = -1 + 36 = 35 \)
In simple words: Put the given values into the cubed difference equation and solve for the cubed terms. Watch your signs when multiplying negative numbers.

Exam Tip: Pay extra attention to the sign when calculating \( -3 \times (-12) \times (-1) \), which results in \( -36 \).

Question 21. If \( m - n = -2 \) and \( m^3 - n^3 = -26 \), find the value of \( mn \).
Answer: We are given \( m - n = -2 \) and \( m^3 - n^3 = -26 \).
Let us write down the identity for the cubed difference:
\( (m - n)^3 = m^3 - n^3 - 3mn(m - n) \)
Substitute the known values into this equation:
\( \implies (-2)^3 = -26 - 3mn(-2) \)
\( \implies -8 = -26 + 6mn \)
Rearranging the equation to solve for \( 6mn \):
\( \implies 6mn = -8 + 26 = 18 \)
Dividing by 6:
\( \implies mn = 3 \)
In simple words: Plug the given values into the standard cubic identity. Rearrange the terms carefully to find the value of the product \( mn \).

Exam Tip: Ensure you evaluate \( (-2)^3 \) as \( -8 \), as negative bases raised to odd powers remain negative.

Question 22. If \( 2a - 3b = 10 \) and \( ab = 16 \), find the value of \( 8a^3 - 27b^3 \).
Answer: We are given \( 2a - 3b = 10 \) and \( ab = 16 \).
Using the identity for cubed difference \( (x - y)^3 = x^3 - y^3 - 3xy(x - y) \):
\( (2a - 3b)^3 = (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b) \)
\( \implies (2a - 3b)^3 = 8a^3 - 27b^3 - 18ab(2a - 3b) \)
Substituting the known values into the equation:
\( \implies (10)^3 = 8a^3 - 27b^3 - 18(16)(10) \)
\( \implies 1000 = 8a^3 - 27b^3 - 2880 \)
Rearranging the terms to isolate \( 8a^3 - 27b^3 \):
\( \implies 8a^3 - 27b^3 = 1000 + 2880 = 3880 \)
In simple words: Expand the cubed difference of \( (2a - 3b) \). Substitute the given values and perform the calculations to solve for the cubed terms.

Exam Tip: Be sure to expand the cross product term as \( 3 \times 2a \times 3b = 18ab \), rather than getting confused with other multipliers.

Question 23. If \( x + 2y = 5 \), prove that \( x^3 + 8y^3 + 30xy = 125 \).
Answer: We start with the given relationship:
\( x + 2y = 5 \)
Let us cube both sides of this equation:
\( (x + 2y)^3 = 5^3 \)
Using the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \), we expand the left side:
\( \implies (x)^3 + (2y)^3 + 3(x)(2y)(x + 2y) = 125 \)
\( \implies x^3 + 8y^3 + 6xy(x + 2y) = 125 \)
Substituting the value \( x + 2y = 5 \) back into the expression:
\( \implies x^3 + 8y^3 + 6xy(5) = 125 \)
\( \implies x^3 + 8y^3 + 30xy = 125 \)
In simple words: Cube both sides of the equation. Expand using the algebraic formula and plug in the initial value to obtain the final proof.

Exam Tip: Substituting the initial equation value back into the middle of the expansion is the key trick to complete this proof smoothly.

Question 24A. Simplify: \( (4x + 5y)^2 + (4x - 5y)^2 \)
Answer: Let us expand both terms using standard identities:
\( (4x + 5y)^2 + (4x - 5y)^2 \)
\( = [(4x)^2 + (5y)^2 + 2(4x)(5y)] + [(4x)^2 + (5y)^2 - 2(4x)(5y)] \)
\( = 16x^2 + 25y^2 + 40xy + 16x^2 + 25y^2 - 40xy \)
Combining the like terms, the \( 40xy \) and \( -40xy \) terms cancel out:
\( = 32x^2 + 50y^2 \)
In simple words: Expand each squared term fully. The middle terms with opposite signs will cancel out, leaving only the squared parts.

Exam Tip: Notice that this matches the pattern \( (A+B)^2 + (A-B)^2 = 2(A^2 + B^2) \), which you can use to check your answer quickly.

Question 24B. Simplify: \( (7a + 5b)^2 - (7a - 5b)^2 \)
Answer: We expand both squared brackets using algebraic identities:
\( (7a + 5b)^2 - (7a - 5b)^2 \)
\( = [(7a)^2 + (5b)^2 + 2(7a)(5b)] - [(7a)^2 + (5b)^2 - 2(7a)(5b)] \)
\( = [49a^2 + 25b^2 + 70ab] - [49a^2 + 25b^2 - 70ab] \)`
Distribute the negative sign through the second bracket:
\( = 49a^2 + 25b^2 + 70ab - 49a^2 - 25b^2 + 70ab \)
The squared terms cancel each other out:
\( = 70ab + 70ab \)
\( = 140ab \)
In simple words: Expand both terms and subtract them. The squares cancel out, and the middle terms add up to give the final product.

Exam Tip: Always be careful with the negative sign when expanding and distributing across brackets to avoid sign mistakes.

Question 24C. Simplify: \( (a + b)^3 + (a - b)^3 \)
Answer: Let us expand both cubed expressions using the cubic identities:
\( (a + b)^3 + (a - b)^3 \)
\( = (a^3 + b^3 + 3a^2b + 3ab^2) + (a^3 - b^3 - 3a^2b + 3ab^2) \)
Now, we group and combine the like terms:
\( = (a^3 + a^3) + (b^3 - b^3) + (3a^2b - 3a^2b) + (3ab^2 + 3ab^2) \)
\( = 2a^3 + 6ab^2 \)
In simple words: Write out the expansion for both cubed brackets. The terms \( b^3 \) and \( 3a^2b \) cancel out because they have opposite signs, leaving the final expression.

Exam Tip: Remember that \( (a-b)^3 \) has alternating signs: \( a^3 - 3a^2b + 3ab^2 - b^3 \).

Question 24D. Simplify: \( \left(a - \frac{1}{a}\right)^2 + \left(a + \frac{1}{a}\right)^2 \)
Answer: We expand both squared binomial terms:
\( \left(a - \frac{1}{a}\right)^2 + \left(a + \frac{1}{a}\right)^2 \)
\( = [(a)^2 + \left(\frac{1}{a}\right)^2 - 2(a)\left(\frac{1}{a}\right)] + [(a)^2 + \left(\frac{1}{a}\right)^2 + 2(a)\left(\frac{1}{a}\right)] \)
\( = a^2 + \frac{1}{a^2} - 2 + a^2 + \frac{1}{a^2} + 2 \)
Combining the terms, the constant values \( -2 \) and \( +2 \) cancel out:
\( = 2a^2 + \frac{2}{a^2} \)
In simple words: Square both parts separately. The constant middle terms will cancel each other out, leaving twice the sum of the squares.

Exam Tip: Check that the cross-multiplication term simplifies to a constant 2 before completing the addition.

Question 24E. Simplify: \( (x + y - z)^2 + (x - y + z)^2 \)
Answer: We expand both terms using the trinomial squaring formula \( (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \):
\( (x + y - z)^2 + (x - y + z)^2 \)
\( = [x^2 + y^2 + z^2 + 2xy - 2yz - 2zx] + [x^2 + y^2 + z^2 - 2xy - 2yz + 2zx] \)
Group and combine the like terms together:
\( = 2x^2 + 2y^2 + 2z^2 + (2xy - 2xy) + (-2yz - 2yz) + (-2zx + 2zx) \)
\( = 2x^2 + 2y^2 + 2z^2 - 4yz \)
In simple words: Expand each bracket with three terms. Many of the cross-product terms cancel out, leaving a simplified expression.

Exam Tip: Keep careful track of negative signs when finding the cross products for the trinomial expansions.

Question 24F. Simplify: \( \left(a + \frac{1}{a}\right)^3 - \left(a - \frac{1}{a}\right)^3 \)
Answer: We expand the cubed brackets using standard algebraic identities:
\( \left(a + \frac{1}{a}\right)^3 - \left(a - \frac{1}{a}\right)^3 \)
\( = [(a)^3 + \left(\frac{1}{a}\right)^3 + 3(a)\left(\frac{1}{a}\right)\left(a + \frac{1}{a}\right)] - [(a)^3 - \left(\frac{1}{a}\right)^3 - 3(a)\left(\frac{1}{a}\right)\left(a - \frac{1}{a}\right)] \)
\( = [a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right)] - [a^3 - \frac{1}{a^3} - 3\left(a - \frac{1}{a}\right)] \)
\( = [a^3 + \frac{1}{a^3} + 3a + \frac{3}{a}] - [a^3 - \frac{1}{a^3} - 3a + \frac{3}{a}] \)
Now, distribute the negative sign through the second bracket:
\( = a^3 + \frac{1}{a^3} + 3a + \frac{3}{a} - a^3 + \frac{1}{a^3} + 3a - \frac{3}{a} \)
Combining like terms:
\( = \frac{2}{a^3} + 6a \)
In simple words: Cube both parts and subtract the second from the first. Many terms cancel, and the remaining ones add up to the final result.

Exam Tip: When distributing the minus sign to the second bracket, make sure every term inside changes its sign correctly.

Question 24G. Simplify: \( (2x + y)(4x^2 - 2xy + y^2) \)
Answer: We multiply each term of the first bracket by the second expression:
\( (2x + y)(4x^2 - 2xy + y^2) \)
\( = 2x(4x^2 - 2xy + y^2) + y(4x^2 - 2xy + y^2) \)
\( = 8x^3 - 4x^2y + 2xy^2 + 4x^2y - 2xy^2 + y^3 \)
Combine the like terms to simplify:
\( = 8x^3 + y^3 \)
In simple words: Multiply each term in the first bracket by the second bracket. The middle terms will cancel out, leaving the sum of cubes.

Exam Tip: This multiplication matches the standard identity \( (A+B)(A^2 - AB + B^2) = A^3 + B^3 \), where \( A = 2x \) and \( B = y \).

Question 24H. Simplify: \( \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right) \)
Answer: We expand the product by distributing the terms:
\( \left(x - \frac{1}{x}\right)\left(x^2 + 1 + \frac{1}{x^2}\right) \)
\( = x\left(x^2 + 1 + \frac{1}{x^2}\right) - \frac{1}{x}\left(x^2 + 1 + \frac{1}{x^2}\right) \)
\( = x^3 + x + \frac{1}{x} - x - \frac{1}{x} - \frac{1}{x^3} \)
The intermediate terms cancel out:
\( = x^3 - \frac{1}{x^3} \)
In simple words: Multiply the terms in the brackets. The intermediate terms cancel out, leaving the difference of cubes.

Exam Tip: This is the expansion of the identity \( (A-B)(A^2 + AB + B^2) = A^3 - B^3 \), where \( A = x \) and \( B = \frac{1}{x} \).

Question 24I. Simplify: \( (x + 2y + 3z)(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) \)
Answer: We expand the expression term-by-term:
\( (x + 2y + 3z)(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) \)
\( = x(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) + 2y(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) + 3z(x^2 + 4y^2 + 9z^2 - 2xy - 6yz - 3zx) \)
Expanding all terms:
\( = x^3 + 4xy^2 + 9xz^2 - 2x^2y - 6xyz - 3zx^2 + 2yx^2 + 8y^3 + 18yz^2 - 4xy^2 - 12y^2z - 6xyz + 3zx^2 + 12zy^2 + 27z^3 - 6xyz - 18yz^2 - 9z^2x \)
Combining and canceling the terms that sum to zero:
\( = x^3 + 8y^3 + 27z^3 - 18xyz \)
In simple words: Multiply the terms of the first expression by each term of the second. After simplifying and canceling the opposite terms, we are left with the sum of cubes minus three times the product.

Exam Tip: This is a direct application of the identity \( (A+B+C)(A^2+B^2+C^2-AB-BC-CA) = A^3+B^3+C^3-3ABC \).

Question 24J. Simplify: \( (1+x)(1-x)(1-x+x^2)(1+x+x^2) \)
Answer: Let us rearrange the terms to group them into standard product identities:
\( (1+x)(1-x)(x^2 - x + 1)(x^2 + x + 1) \)
Group as \( (1-x)(1+x+x^2) \) and \( (1+x)(1-x+x^2) \) to use cubic identities:
\( = [(1-x)(x^2 + x + 1)][(1+x)(x^2 - x + 1)] \)
Using the identities \( (a-b)(a^2+ab+b^2) = a^3-b^3 \) and \( (a+b)(a^2-ab+b^2) = a^3+b^3 \):
\( = (1 - x^3)(1 + x^3) \)
Applying the difference of squares identity \( (a-b)(a+b) = a^2 - b^2 \):
\( = 1^2 - (x^3)^2 \)
\( = 1 - x^6 \)
In simple words: Pair the factors together to form sums and differences of cubes. Multiply the resulting two terms to get the final simplified expression.

Exam Tip: Grouping terms strategically makes solving complex-looking expressions much simpler and reduces computation errors.

Question 24K. Simplify: \( (3a + 2b - c)(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca) \)
Answer: We expand by distributing each term from the first bracket through the second bracket:
\( (3a + 2b - c)(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca) \)
\( = 3a(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca) + 2b(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca) - c(9a^2 + 4b^2 + c^2 - 6ab + 2bc + 3ca) \)
Multiplying term by term:
\( = 27a^3 + 12ab^2 + 3ac^2 - 18a^2b + 6abc + 9a^2c + 18a^2b + 8b^3 + 2bc^2 - 12ab^2 + 4b^2c + 6abc - 9a^2c - 4b^2c - c^3 + 6abc - 2bc^2 - 3ac^2 \)
Combining and canceling the terms that add up to zero:
\( = 27a^3 + 8b^3 - c^3 + 18abc \)
In simple words: Multiply each term from the first group with the second group. Most of the cross terms cancel out, leaving the sum of cubes and the product term.

Exam Tip: This is the expansion of the identity \( (A+B+C)(A^2+B^2+C^2-AB-BC-CA) = A^3+B^3+C^3-3ABC \) with \( A = 3a \), \( B = 2b \), and \( C = -c \).

Question 24L. Simplify: \( (3x + 5y + 2z)(3x - 5y + 2z) \)
Answer: We rearrange the terms to group them and use the difference of squares identity:
\( (3x + 5y + 2z)(3x - 5y + 2z) \)
\( = [(3x + 2z) + 5y][(3x + 2z) - 5y] \)
Applying the identity \( (A + B)(A - B) = A^2 - B^2 \):
\( = (3x + 2z)^2 - (5y)^2 \)
Expanding the squared terms:
\( = [9x^2 + 2(3x)(2z) + 4z^2] - 25y^2 \)
\( = 9x^2 + 12xz + 4z^2 - 25y^2 \)
Rearranging the terms in standard order:
\( = 9x^2 - 25y^2 + 4z^2 + 12xz \)
In simple words: Group \( 3x + 2z \) as one term and apply the difference of squares formula. Expand the remaining square to find the final result.

Exam Tip: Grouping terms that have the same sign makes it easy to use the difference of squares identity directly.

Question 24M. Simplify: \( (2x - 4y + 7)(2x + 4y + 7) \)
Answer: Rearranging the terms to group them for the difference of squares identity:
\( (2x - 4y + 7)(2x + 4y + 7) \)
\( = [(2x + 7) - 4y][(2x + 7) + 4y] \)
Using the identity \( (A - B)(A + B) = A^2 - B^2 \):
\( = (2x + 7)^2 - (4y)^2 \)
Expanding the squared terms:
\( = [4x^2 + 2(2x)(7) + 7^2] - 16y^2 \)
\( = 4x^2 + 28x + 49 - 16y^2 \)
Ordering the terms with variables first:
\( = 4x^2 - 16y^2 + 28x + 49 \)
In simple words: Group \( 2x + 7 \) together and use the difference of squares formula. Then, expand the perfect square to get the answer.

Exam Tip: Look for identical parts with the same signs to identify your \( A \) term, and parts with opposite signs for your \( B \) term.

Question 24N. Simplify: \( (3a - 7b + 3)(3a - 7b + 5) \)
Answer: We can distribute the terms from the first expression across the second expression:
\( (3a - 7b + 3)(3a - 7b + 5) \)
\( = 3a(3a - 7b + 5) - 7b(3a - 7b + 5) + 3(3a - 7b + 5) \)
Multiply term by term:
\( = 9a^2 - 21ab + 15a - 21ab + 49b^2 - 35b + 9a - 21b + 15 \)
Group and combine the like terms together:
\( = 9a^2 - (21ab + 21ab) + (15a + 9a) + 49b^2 - (35b + 21b) + 15 \)
\( = 9a^2 - 42ab + 24a + 49b^2 - 56b + 15 \)
In simple words: Multiply every term in the first bracket by each term in the second bracket, then group similar terms together to simplify.

Exam Tip: Be methodical and write down every single term to avoid missing any cross multiplication during the expansion.

Question 24O. Simplify: \( (4m - 5n - 8)(4m - 5n + 5) \)
Answer: We expand the expression by distributing each term of the first bracket:
\( (4m - 5n - 8)(4m - 5n + 5) \)
\( = 4m(4m - 5n + 5) - 5n(4m - 5n + 5) - 8(4m - 5n + 5) \)
Multiplying each term:
\( = 16m^2 - 20mn + 20m - 20mn + 25n^2 - 25n - 32m + 40n - 40 \)
Now, group and combine the like terms:
\( = 16m^2 + 25n^2 - (20mn + 20mn) + (20m - 32m) + (-25n + 40n) - 40 \)
\( = 16m^2 + 25n^2 - 40mn - 12m + 15n - 40 \)
In simple words: Distribute the terms to multiply the brackets fully, then simplify by adding and subtracting like terms.

Exam Tip: Triple check your addition and subtraction of like terms, such as \( 20m - 32m = -12m \).

Question 25. Evaluate using algebraic identities:
(i) \( (3.29)^3 + (6.71)^3 \)
(ii) \( (5.45)^3 + (3.55)^3 \)
(iii) \( (8.12)^3 - (3.12)^3 \)
(iv) \( 7.16 \times 7.16 + 2.16 \times 7.16 + 2.16 \times 2.16 \)
(v) \( 1.81 \times 1.81 - 1.81 \times 2.19 + 2.19 \times 2.19 \)
Answer:
(i) Using the identity \( x^3 + y^3 = (x+y)^3 - 3xy(x+y) \) where \( x = 3.29 \) and \( y = 6.71 \):
\( (3.29)^3 + (6.71)^3 \)
\( = (3.29 + 6.71)^3 - 3(3.29)(6.71)(3.29 + 6.71) \)
\( = (10)^3 - 3(3.29)(6.71)(10) \)
Using the identity \( (a-b)(a+b) = a^2 - b^2 \) for the terms \( 3.29 = 5 - 1.71 \) and \( 6.71 = 5 + 1.71 \):
\( = 1000 - 30(5 - 1.71)(5 + 1.71) \)
\( = 1000 - 30[(5)^2 - (1.71)^2] \)
\( = 1000 - 30(25 - 2.9241) \)
\( = 1000 - 30 \times 22.0759 \)
\( = 1000 - 662.277 \)
\( = 337.723 \)

(ii) Using the identity \( x^3 + y^3 = (x+y)^3 - 3xy(x+y) \) where \( x = 5.45 \) and \( y = 3.55 \):
\( (5.45)^3 + (3.55)^3 \)
\( = (5.45 + 3.55)^3 - 3(5.45)(3.55)(5.45 + 3.55) \)
\( = (9)^3 - 3(5.45)(3.55)(9) \)
Expressing \( 5.45 = 4.5 + 0.95 \) and \( 3.55 = 4.5 - 0.95 \) to use the difference of squares identity:
\( = 729 - 27(4.5 + 0.95)(4.5 - 0.95) \)
\( = 729 - 27[(4.5)^2 - (0.95)^2] \)
\( = 729 - 27(20.25 - 0.9025) \)
\( = 729 - 27 \times 19.3475 \)
\( = 729 - 522.3825 \)
\( = 206.6175 \)

(iii) Using the identity \( x^3 - y^3 = (x-y)^3 + 3xy(x-y) \) where \( x = 8.12 \) and \( y = 3.12 \):
\( (8.12)^3 - (3.12)^3 \)
\( = (8.12 - 3.12)^3 + 3(8.12)(3.12)(8.12 - 3.12) \)
\( = 5^3 + 3(8.12)(3.12) \times 5 \)
\( = 125 + 15 \times 25.3344 \)
\( = 125 + 380.016 \)
\( = 505.016 \)

(iv) Simplifying \( 7.16 \times 7.16 + 2.16 \times 7.16 + 2.16 \times 2.16 \):
\( = (7.16)^2 + (2.16)(7.16) + (2.16)^2 \)
Add and subtract \( (2.16)(7.16) \) to construct a perfect square:
\( = (7.16)^2 + (2.16)(7.16) + (2.16)^2 + (2.16)(7.16) - (2.16)(7.16) \)
\( = (7.16)^2 + 2(2.16)(7.16) + (2.16)^2 - (2.16)(7.16) \)
\( = (7.16 + 2.16)^2 - (2.16)(7.16) \)
\( = (9.32)^2 - 15.4656 \)
\( = 86.8624 - 15.4656 \)
\( = 71.3968 \)

(v) Simplifying \( 1.81 \times 1.81 - 1.81 \times 2.19 + 2.19 \times 2.19 \):
\( = (1.81)^2 - (1.81 \times 2.19) + (2.19)^2 \)
Add and subtract \( (1.81 \times 2.19) \) to form a perfect square:
\( = (1.81)^2 - (1.81 \times 2.19) + (2.19)^2 - (1.81 \times 2.19) + (1.81 \times 2.19) \)
\( = (1.81)^2 - 2(1.81 \times 2.19) + (2.19)^2 + (1.81 \times 2.19) \)
\( = (1.81 - 2.19)^2 + (2.00 - 0.19)(2.00 + 0.19) \)
\( = (-0.38)^2 + [2^2 - (0.19)^2] \)
\( = 0.1444 + (4 - 0.0361) \)
\( = 0.1444 + 3.9639 \)
\( = 4.1083 \)
In simple words: Use the sum and difference of cubes identities, or perfect square identities, to rewrite each expression. This lets you simplify decimal calculations using easier numbers.

Exam Tip: When evaluating decimal calculations, look for patterns that sum to nice whole numbers (like 10 or 9) to simplify your arithmetic.