Q7 of 65 Page 141

Let us factorise the following algebraic expressions:

a6 + 32a3 – 64

a6 + 32a3 – 64


⇒ a6 + 8a3 – 64 + 24a3


⇒ (a2)3 + (2a) 3 + (- 4)3 - 3×(a2)×(2a)×(- 4) …Equation (i)


We use the identity


a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)


Using the above identity in Equation (i) we get


⇒ (a2)3 + (2a) 3 + (- 4)3 - 3×(a2)×(2a)×(- 4) = (a2 + 2a - 4) (a4 + 4a2 + 16 - 2a3 + 8a + 4a2)


⇒ (a2 + 2a - 4) (a4 + 8a2 + 16 - 2a3 + 8a)


(a2 + 2a - 4) (a4 - 2a3 + 8a2 + 8a + 16)


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