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

Question 1.

Factorize

i. y³ – 27

ii. x³ – 64y³

iii. 27m³ – 216n³

iv. 125y³ – 1

v. \(8 p^{3}-\frac{27}{p^{3}}\)

vi. 343a³ – 512b³

vii. 64x³ – 729y³

viii. \(16 a^{3}-\frac{128}{b^{3}}\)

Solution:

i. y³ – 27

= y³ – (3)³

Here, a = y and b = 3

∴ y³ – 27 = (y – 3)[y² + y(3) + (3)2]

…[∵ a³ – b³ = (a – b) (a² + ab + b²)]

= (y – 3)(y² + 3y + 9)

ii. x³ – 64y³

= x³ – (4y)³

Here, a = x and b = 4y

∴ x³ – 64y³ = (x – 4y)[x² + x(4y) + (4y)²]

…[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (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

∴ 27m³ – 216n³

= 27 {(m – 2n) [m² + m(2n) + (2n)²]}

….[∵ a³ – b³ = (a – b) (a² + ab + b²)]

= 27 (m – 2n)(m² + 2mn + 4n²)

iv. 125y³ – 1

= (5y)³ – 1³

Here, a = 5y and b = 1

∴ 125y³ – 1 = (5y – 1) [(5y)² + (5y)(1) + (1)²]

…[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (5y – 1) (25y² + 5y + 1)

v. \(8 p^{3}-\frac{27}{p^{3}}\)

vi. 343a³ – 512b³

= (7a)³ – (8b)³

Here, A = 7a and B = 8b

∴ 343a³ – 512b³

= (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]

…[∵ A³ – B³ = (A – B)(A² + AB + B²)]

= (7a – 8b) (49a² + 56ab + 64b²)

vii. 64x³ – 729y³

= (4x)³ – (9y)³

Here, a = 4x and b = 9y

∴ 64x³ – 729y³

= (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]

…[∵ a³ – b³ = (a – b)(a² + ab + b²)]

= (4x – 9y) (16x² + 36xy + 81y²)

viii. \(16 a^{3}-\frac{128}{b^{3}}\)

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

Solution:

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)]

…[a³ – b³ = (a – b)(a² + ab + b²)]

= (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]

…[∵ A³ – B³ = (A – B)(A² + AB + B²)]

= (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) …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]

= p³ – p³ – 3p² – 3p – 1

= – 3p² – 3p – 1

v. (3xy – 2ab)³ – (3xy + 2ab)³

Here, A = 3xy – 2ab and B = 3xy + 2ab

∴ (3xy – 2ab)³ – (3xy + 2ab)³

= [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]

…[∵ A³ – B³ = (A – B) (A² + AB + B²)]

= (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