Q2 of 65 Page 141

Let us factorise the following algebraic expressions:

8x3 – y3 + 1 + 6xy

8x3 – y3 + 1 + 6xy


⇒ (2x) 3 + (- y) 3 + 13 + 6xy


⇒ (2x) 3 + (- y) 3 + 13 - 3×(2x)×(- y)×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


⇒ (2x) 3 + (- y) 3 + 13 - 3×(2x)×(- y) = (2x – y + 1) ((2x)2 + y2 + 12 + 2xy + y - 2x)


⇒ (2x) 3 + (- y) 3 + 13 - 3×(2x)×(- y) = (2x – y + 1) (4x2 + y2 + 1 + 2xy + y - 2x)


(2x – y + 1) (4x2 + y2 + 1 + 2xy + y - 2x)


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