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

Maybe there is a trick to it, but good ol' factoring can work here.
It is easy to check (using the identity $a^3=-a-1$) that
$$ (3a^2-2a+2)(3a^2+2a) = -5a-7a^2 $$
Let $\sqrt[3]{-5a-7a^2}+a^2 = y$. Given that $a\in \mathbb{R}$, we know that also $y\in \mathbb{R}$, and we must solve for it.
$$y-a^2 = \sqrt[3]{-5a-7a^2} \\ (y-a^2)^3 = -5a-7a^2 \\ y^3 - 3y^2a^2 + 3ya^4-a^6 = -5a-7a^2 \\ y^3-3y^2a^2+3ya(-a-1)-(-a-1)^2=-5a-7a^2 \\ y^3-3y^2a^2-3ya^2-3ya+6a^2+3a-1=0$$ So many threes, only the $6$ stands out. But if we use $6=3+3$: $$ (y^3-1)+(3a^2-3y^2a^2)+(3a^2+3a-3ya^2-3ya)=0 \\ (y^3-1)-3a^2(y^2-1)-(3a^2+3a)(y-1)=0$$ As one can see, it is possible to factor out $(y-1)$, thereby concluding that $y=1$ is a solution.  
