if it is what soumik is saying then
a^{3}+b^{3}=(a+b)^{3}
hence solutions will be
a=0
b=0
or
a+b=0
solve for x .
(x^{2} + x -3)^{3} + (2x^{2} - x - 1)^{3} = 27(x^{2} - 1)^{3}
This is certainly not the question.
The real eqn is (x^{2}+x-2)^{3}+(2x^{2}-x-1)^{3}=27(x^{2}-1)^{3} & the roots of this turn out to be (1,-1,-2,-12)
if it is what soumik is saying then
a^{3}+b^{3}=(a+b)^{3}
hence solutions will be
a=0
b=0
or
a+b=0
sorry there soumik.
i has noted it wrong. that is what i was thinking.
if it was -2 in place of -3.
then (x^{2}+x-2)^{3}+(2x^{2}-x-1)^{2}=27(x^{2}-1)^{3}
then we can get
{(x-1)(x-2)}^{3} + {(x-1)(2x+1)}^{2} - 27{(x+1)(x-1)}^{3} = 0
clearly x = 1 is a root.
(x-1)(x-2)^{3} + (2x+1)^{2} - 27(x+1)^{3}(x-1) = 0
the roots will be when all the terms are 0.
that is -1, 2 and -1/2.