Q8 of 65 Page 141

Let us factorise the following algebraic expressions:

a6 – 18a3 + 125

a6 – 18a3 + 125


⇒ a6 + 27a3 + 125 - 45a3


⇒ (a2)3 + (3a) 3 + 53 - 45a3


⇒ (a2)3 + (3a) 3 + 53 – 3× a2× 3a× 5 …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 + (3a) 3 + 53 – 3× a2× 3a× 5 = (a2 + 3a + 5) (a4 + 9a2 + 25 - 3a3 - 15a - 5a2)


⇒ (a2)3 + (3a) 3 + 53 – 3× a2× 3a× 5 = (a2 + 3a + 5) (a4 + 4a2 + 25 - 3a3 - 15a)


(a2 + 3a + 5) (a4 - 3a3 + 4a2 - 15a + 25)


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