
2. Solve the following equation for real $x$ :

$$
\left(x^{2}+x-2\right)^{3}+\left(2 x^{2}-x-1\right)^{3}=27\left(x^{2}-1\right)^{3}
$$


Solution: By setting $u=x^{2}+x-2$ and $v=2 x^{2}-x-1$, we observe that the equation reduces to $u^{3}+v^{3}=(u+v)^{3}$. Since $(u+v)^{3}=u^{3}+v^{3}+3 u v(u+v)$, it follows that $u v(u+v)=0$. Hence $u=0$ or $v=0$ or $u+v=0$. Thus we obtain $x^{2}+x-2=0$ or $2 x^{2}-x-1=0$ or $x^{2}-1=0$. Solving each of them we get $x=1,-2$ or $x=1,-1 / 2$ or $x=1,-1$. Thus $x=1$ is a root of multiplicity 3 and the other roots are $-1,-2,-1 / 2$.

(Alternately, it can be seen that $x-1$ is a factor of $x^{2}+x-2,2 x^{2}-x-1$ and $x^{2}-1$. Thus we can write the equation in the form

$$
(x-1)^{3}(x+2)^{3}+(x-1)^{3}(2 x+1)^{3}=27(x-1)^{3}(x+1)^{3}
$$

Thus it is sufficient to solve the cubic equation

$$
(x+2)^{3}+(2 x+1)^{3}=27(x+1)^{3}
$$

This can be solved as earlier or expanding every thing and simplifying the relation.)
