Question

"Question 8 Factorise each of the following:
(i) 8a^3 + b^3 + 12a^2b + 6ab^2
(ii) 8a^3 - b^3 - 12a^2b + 6ab^2
(iii) 27 - 125a^3 - 135a + 225a^2
(iv) 64a^3 - 27b^3 - 144a^2b + 108ab^2
(v) 27p^3 - 1/216 - 9p^2/2 + p/4

Class 9 - Math - Polynomials Page 49"

Answer 1

Identity:

An identity is an equality which is true for all values of a variable in the equality.

(a + b)³ = a³+ b³+ 3ab(a + b)

In an identity the right hand side expression is called expanded form of the left hand side expression.

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Solution:

(i) 8a³ + b³+ 12a²b + 6ab²

Using identity,

(a + b)³ = a³ + b³ + 3a²b + 3ab²

8a³ + b³+ 12a²b + 6ab²

 

We can rewrite a given expression as

= (2a)³ + b³ + 3(2a)²b + 3(2a)(b)²

= (2a + b)³

= (2a + b)(2a + b)(2a + b)

 

(ii) 8a³ – b³ – 12a²b + 6ab²

Using identity,

(a – b)³ = a³– b³ – 3a²b + 3ab²

8a³ – b³– 12a²b + 6ab²

We can rewrite the given expression as

= (2a)³ – b³ – 3(2a)²b + 3(2a)(b)²

= (2a – b)³

= (2a – b)(2a – b)(2a – b)

 

(iii) 27 – 125a³ – 135a + 225a²

Using identity,

 (a – b)³ = a³ – b³ – 3a²b + 3ab²

27 – 125a³ – 135a + 225a²

= 3³– (5a)³ – 3(3)²(5a) + 3(3)(5a)²

= (3 – 5a)³

= (3 – 5a)(3 – 5a)(3 – 5a)

 

(iv) 64a³ – 27b³ – 144a²b + 108ab²

Using identity,

 (a – b)³ = a³ – b³ – 3a²b + 3ab²

64a³ – 27b³ – 144a²b + 108ab²

 

We can rewrite the  given expression as

= (4a)³ – (3b)³ – 3(4a)²(3b) + 3(4a)(3b)²

= (4a – 3b)³

=(4a–3b)(4a–3b)(4a –3b)

 

(v) 27p³ – 1/216 – 9/2 p²+ 1/4 p

 

 Using identity,

 (a – b)³ = a³ – b³ – 3a²b + 3ab²

 27p³ – 1/216 – 9/2 p² + 1/4 p

 

We can rewrite the given expression as

= (3p)³-(1/6)³-3(3p)²(1/6) + 3(3p)(1/6)²

= (3p – 1/6)³

= (3p – 1/6) (3p – 1/6) (3p – 1/6)

 

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 Hope this will help you....

Answer 2

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