Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s

The real root $a$ of $x^3+x+1$ belongs to the interval $\left(-1,-\frac{1}{2}\right)$ and in order to check that
$$ \sqrt[3]{(3a^2-2a+2)(3a^2+2a)}=1-a^2 \tag{1}$$
holds it is enough to check that
$$ (3a^2-2a+2)(3a^2+2a)=1-3a^2+3a^4+a^6 \tag{2}$$
holds, or that 
$$  (1+a+a^3)(-1+5a+a^3) = 0 \tag{3} $$
holds, but that is trivial.
