Q3 of 65 Page 141

Let us factorise the following algebraic expressions:

8a3 – 27b3 – 1 – 18ab

8a3 – 27b3 – 1 – 18ab


⇒ (2a) 3 + (- 3b) 3 + (- 1)3 - 18ab


⇒ (2a) 3 + (- 3b) 3 + (- 1)3 - 3×(2a)×(- 3b)×(- 1) …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


⇒ (2a) 3 + (- 3b) 3 + (- 1)3 - 3×(2a)×(- 3b)×(- 1) = (2a - 3b - + 1) (4a2 + 9b2 + 1 + 6ab - 3b - 2a)


(2a - 3b - 1) (4a2 + 9b2 + 1 - 6ab + 3b + 2a)


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