Question

Factorise: 

i. x³ + 64y³ 

ii. 125p³ + q³

iii. 125k³ + 27m³

iv. 2l³ + 432m³ 

v. 24a³ + 81b³

vi. y³ + (1/8y³)

vii. a³ + (8/a³)

viii. 1 + (q³/125)

Answer 1

i. 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)

ii. 125p³ + q³ 

= (5p)³ + q³ 

Here, a = 5p and b = q 

∴ 125p³ + q³ = (5p + q)[(5p)² – (5p)(q) + q²] …[∵ a³ + b³ = (a + b)(a² – ab + b²)] 

= (5p + q)(25p² – 5pq + q²)

iii. 125k³ + 27m³ 

= (5k)³ + (3m)³ 

Here, a = 5k and b = 3m 

∴ 125k³ + 27m³ = (5k + 3m) [(5k)² – (5k)(3m) + (3m)²] …[∵ a³ + b³ = (a + b)(a² – ab + b²)] 

= (5k + 3m)(25k² – 15km + 9m²) 

iv. 2l³ + 432m³ 

= 2 (l³ + 216m³) … [Taking out the common factor 2] 

= 2[l³ + (6m)³] 

Here, a = l and b = 6m 

2l³ + 432m³ = 2{(l + 6m)[l² – l(6m) + (6m)²]} …[∵ a³ + b³ = (a + b)(a² – ab + b²)] 

= 2(l + 6m)(l² – 6lm + 36m²) 

v. 24a³ + 81b³ …[Taking out the common factor 3] 

= 3[(2a)³ + (3b)³] 

Here, A = 2a and B = 3b 

∴ 24a³ + 81b³ = 3{(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]} …[∵ A³ + B³ 

= (A + B) (A² – AB + B²)] = 3(2a + 3b)(4a² – 6ab + 9b²)

vi. y³ + (1/8y³)

vii. a³ + (8/a³)

viii. 1 + (q³/125)