IIT JEE , AIEEE Forum - Mathematics , Algebra

sentinel

Solve the following equation for real x?

(x^2 + x - 2)^3 + (2x^2 - x - 1)^3 = 27(x^2 - 1)^3

hsbhatt

This is equivalent to

a^3+b^3 = (a+b)^3 where

a = x^2 + x -2 and b = 2x^2 - 2x - 1

Hence, we have $(a+b)^3 - (a^3+b^3) = 0 \Rightarrow 3ab(a+b) = 0$

Hence either a=0 or b=0 or a+b = 0

This gives the solutions $x = \pm 1; x = -\frac{1}{2}; x = -2$

Decoder

here is another way...(though not tht touch of exellence)...

cancel (x-1)^3...u get x=1 as soln.

now u observe tht a^3 + b^3 -c^3 = 0...

also observe tht a+b+c=0...wic simply means tht a,b,c r 0..

providing us solns. -2 , -1 and -1/2 ..

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