## Chapter 6 Factorisation of Algebraic Expressions Set 6.2

Question 1.

Factorise:

i. x³ + 64y³

ii. 125p³ + q³

iii. 125k³ + 27m³

iv. 2l³ + 432m³

v. 24a³ + 81b³

Solution:

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)