Question

Factorize 

i. y³ – 27 

ii. x³ – 64y³

iii. 27m³ – 216n³ 

iv. 125y³ – 1 

v. 8p³ - (27/p³)

vi. 343a³ – 512b³ 

vii. 64x³ – 729y³

viii. 16a³ - (128/b³)

Answer 1

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. 8p³ - (27/p³)

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. 16a³ - (128/b³)