11This is certainly not the question.
The real eqn is (x2+x-2)3+(2x2-x-1)3=27(x2-1)3 & the roots of this turn out to be (1,-1,-2,-12)
solve for x .
(x2 + x -3)3 + (2x2 - x - 1)3 = 27(x2 - 1)3
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3 Answers
111if it is what soumik is saying then
a3+b3=(a+b)3
hence solutions will be
a=0
b=0
or
a+b=0
3sorry there soumik.
i has noted it wrong. that is what i was thinking.
if it was -2 in place of -3.
then (x2+x-2)3+(2x2-x-1)2=27(x2-1)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.