Maharashtra Board Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions Set 6.3 Solutions

Question 1. Factorize
(i) y³ – 27
(ii) x³ – 64y³
(iii) 27m³ – 216n³
(iv) 125y³ – 1
(v) 8p³ - \( \frac{27}{p^3} \)
(vi) 343a³-512b³
(vii) 64x³ – 729y³
(viii) 16a³ - \( \frac{128}{b^3} \)
Answer:
(i) y³ – 27
= y³ – (3)³
Here, a = y and b = 3
\( \therefore \) y³ – 27 = (y - 3)[y² + y(3) + (3)²]
\[ \therefore a^3 - b^3 = (a - b) (a^2 + ab + b^2) \] = (y - 3)(y² + 3y + 9)

(ii) x³ – 64y³
= x³ – (4y)³
Here, a = x and b = 4y
\( \therefore \) x³ – 64y³ = (x – 4y)[x² + x(4y) + (4y)²]
\[ \therefore a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] = (x – 4y)(x² + 4xy + 16y²)

(iii) 27m³ – 216n³
= 27 (m³ – 8n³)
[Taking out the common factor 27]
= 27 [m³ – (2n)³]
Here, a = m and b = 2n
\( \therefore \) 27m³ – 216n³
= 27 {(m – 2n) [m² + m(2n) + (2n)²]}
\[ \therefore a^3 - b^3 = (a - b) (a^2 + ab + b^2) \] = 27 (m – 2n)(m² + 2mn + 4n²)

(iv) 125y³ – 1
= (5y)³ – 1³
Here, a = 5y and b = 1
\( \therefore \) 125y³ – 1 = (5y – 1) [(5y)² + (5y)(1) + (1)²]
\[ \therefore a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] = (5y – 1) (25y² + 5y + 1)

(v) 8p³ - \( \frac{27}{p^3} \)
= \((2p)^3 - \left(\frac{3}{p}\right)^3\)
Here, a = 2p and b = \( \frac{3}{p} \)
\( \therefore \) \( 8p^3 - \frac{27}{p^3} \)
= \( \left(2p - \frac{3}{p}\right) \left[(2p)^2 + (2p)\left(\frac{3}{p}\right) + \left(\frac{3}{p}\right)^2\right] \)
\[ \therefore a^3 - b^3 = (a - b) (a^2 + ab + b^2) \]
= \( \left(2p - \frac{3}{p}\right) \left(4p^2 + 6 + \frac{9}{p^2}\right) \)

(vi) 343a³-512b³
= (7a)³ – (8b)³
Here, A = 7a and B = 8b
\( \therefore \) 343a³ – 512b³
= (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]
\[ \therefore A^3 - B^3 = (A - B)(A^2 + AB + B^2) \] = (7a – 8b) (49a² + 56ab + 64b²)

(vii) 64x³ – 729y³
= (4x)³ – (9y)³
Here, a = 4x and b = 9y
\( \therefore \) 64x³ – 729y³
= (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]
\[ \therefore a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] = (4x – 9y) (16x² + 36xy + 81y²)

(viii) 16a³ - \( \frac{128}{b^3} \)
= \( 16 \left(a^3 - \frac{8}{b^3}\right) \)
[Taking out the common factor 16]
= \( 16 \left[a^3 - \left(\frac{2}{b}\right)^3\right] \)
Here, A = a and B = \( \frac{2}{b} \)
\( \therefore \) \( 16a^3 - \frac{128}{b^3} \)
= \( 16 \left[ \left(a - \frac{2}{b}\right) \left(a^2 + a\left(\frac{2}{b}\right) + \left(\frac{2}{b}\right)^2\right) \right] \)
\[ \therefore A^3 - B^3 = (A - B) (A^2 + AB + B^2) \]
= \( 16 \left(a - \frac{2}{b}\right) \left(a^2 + \frac{2a}{b} + \frac{4}{b^2}\right) \)
In simple words: This question requires factoring expressions using the difference of cubes formula, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). For some parts, a common factor needs to be taken out first before applying the formula.

🎯 Exam Tip: Remember the difference of cubes formula and practice identifying perfect cubes. Watch out for common factors, as factoring them out simplifies the problem significantly.

Question 2. Simplify:
(i) (x + y)³ – (x - y)³
(ii) (3a + 5b)³ – (3a – 5b)³
(iii) (a + b)³ – a³ – b³
(iv) p³ – (p + 1)³
(v) (3xy – 2ab)³ – (3xy + 2ab)³
Answer:
(i) (x + y)³ – (x - y)³
Here, a = x + y and b = x - y
(x + y)³ – (x - y)³
= [(x + y) – (x - y)] [(x + y)² + (x + y) (x - y) + (x - y)²]
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
= (x + y - x + y) [(x² + 2xy + y²) + (x² - y²) + (x² – 2xy + y²)]
= 2y(x² + x² + x² + 2xy - 2xy + y² – y² + y²)
= 2y (3x² + y²)
= 6x²y + 2y³

(ii) (3a + 5b)³ – (3a – 5b)³
Here, A = 3a + 5b and B = 3a – 5b
= [(3a + 5b) – (3a – 5b)] [(3a + 5b)² + (3a + 5b) (3a – 5b) + (3a – 5b)²]
\[ \therefore A^3 - B^3 = (A - B)(A^2 + AB + B^2) \]
= (3a + 5b – 3a + 5b) [(9a² + 30ab + 25b²) + (9a² – 25b²) + (9a² – 30ab + 25b²)]
= 10b (9a² + 9a² + 9a² + 30ab – 30ab + 25b² – 25b² + 25b²)
= 10b (27a² + 25b²)
= 270a²b + 250b³

(iii) (a + b)³ – a³ – b³
= a³ + 3a²b + 3ab² + b³ – a³ – b³
= 3a²b + 3ab²

(iv) p³ – (p + 1)³
= p³ – (p³ + 3p² + 3p + 1) \[ \therefore (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \]
= p³ – p³ – 3p² – 3p – 1
= -3p² – 3p – 1

(v) (3xy – 2ab)³ – (3xy + 2ab)³
Here, A = 3xy – 2ab and B = 3xy + 2ab
\( \therefore \) (3xy – 2ab)³ – (3xy + 2ab)³
= [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]
\[ \therefore A^3 - B^3 = (A - B) (A^2 + AB + B^2) \]
= (3xy – 2ab – 3xy – 2ab) [(9x²y² – 12xyab + 4a²b²) + (9x²y² – 4a²b²) + (9x²y² + 12xyab + 4a²b²)]
= (- 4ab) (9x²y² + 9x²y² + 9x²y² – 12xyab + 12xyab + 4a²b² - 4a²b² + 4a²b²)
= (- 4ab) (27 xy² + 4a²b²)
= -108x²y²ab – 16a³b³
In simple words: This question involves simplifying algebraic expressions, primarily by expanding cubes and using the difference of cubes formula, \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), and then combining like terms.

🎯 Exam Tip: Be careful with signs when expanding and combining terms. Practice the formulas for sum and difference of cubes, as well as the expansion of binomials like \((a+b)^3\).