Factorise each of the following:
i) 8a³ + b³ + 12a²b + 6ab²
ii) 8a³ - b³ - 12a²b + 6ab²
iii) 27 - 125a³ - 135a + 225a²
iv) 64a³ - 27b³ - 144a²b + 108ab²
v) 27p³ - 1/216 - (9/2)p² + (1/4)p

Solution:

We will use the following Algebraic identities to solve the following:

(x + y)³ = x³ + y³ + 3x²y + 3xy²

(x - y)³ = x³ - y³ - 3x²y + 3xy²

i) 8a³ + b³ + 12a²b + 6ab²

This can be re-written as: (2a)³ + (b)³ + 3(2a)²(b) + 3(2a)(b)²

which is of the form (x + y)³ = x³ + y³ + 3x²y + 3xy²

Here, x = 2a, y = b

Hence 8a³ + b³ +12a²b + 6ab² = (2a + b)³

ii) 8a³ - b³ - 12a²b + 6ab²

This can be re-written as: (2a)³ - (b)³ - 3(2a)²(b) + 3(2a)(b)²

which is of the form (x - y)³ = x³ - y³ - 3x²y + 3xy²

Here, x = 2a, y = b

Hence 8a³ - b³ - 12a²b + 6ab² = (2a - b)³

iii) 27 - 125a³ - 135a + 225a²

This can be re-written as: (3)³ - (5a)³ - 3(3)²(5a) + 3(3)(5a)²

which is of the form (x - y)³ = x³ - y³ - 3x²y + 3xy²

Here, x = 3, y = 5a

Hence 27 - 125a³ - 135a + 225a² = (3 - 5a)³

iv) 64a³ - 27b³ - 144a²b + 108ab²

This can be re-written as: (4a)³ - (3b)³ - 3(4a)²(3b) + 3(4a)(3b)²

which is of the form (x - y)³ = x³ - y³ - 3x²y + 3xy²

Here, x = 4a, y = 3b

Hence 64a³ - 27b³ - 144a²b + 108ab² = (4a - 3b)³

v) 27p³ - 1/216 - (9/2)p² + (1/4)p

This can be re-written as: (3p)³ - (1/6)³ - 3 × (3p)² × 1/6 + 3 × 3p × (1/6)²

which is of the form (x - y)³ = x³ - y³ - 3x²y + 3xy²

Here, x = 3p, y = 1/6

Hence, 27p³ - 1/216 - (9/2)p² + (1/4)p = (3p - 1/6)³

☛ Check: NCERT Solutions Class 9 Maths Chapter 2

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Video Solution:

Factorise each of the following: i) 8a³ + b³ + 12a²b + 6ab² ii) 8a³ - b³ - 12a²b + 6ab² iii) 27 - 125a³ - 135a + 225a² iv) 64a³ - 27b³ - 144a²b + 108ab² v) 27p³ - 1/216 - (9/2)p² + (1/4)p

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 8

Summary:

The factorized form of the following 8a³ + b³ + 12a²b + 6ab² , 8a³ - b³ - 12a²b + 6ab² , 27 - 125a³ - 135a + 225a² , 64a³ - 27b³ - 144a²b + 108ab² and 27p³ - 1/216 - (9/2)p² + (1/4)p are (2a + b)³, (2a − b)³, (3 − 5a)³, (4a − 3b)³, and (3p − 1/6)³ respectively.

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