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 expansion of what is under square root gives
$$A=9a^4+6a^3-6a^3-4a^2+6a^2+4a=$$
$$a (9a^3+2a+4)=a (9 (-a-1)+2a+4)=$$
$$-a (7a+5)=$$
this must be equal to
$$B=(1-a^2)^3=1-3a^2+3a^4-a^6=$$
$$$$
the difference is
$$B-A=1+4a^2+5a+3a^4-(a+1)^2=$$
$$3a^2+3a+3a^4=3a (a^3+a+1)=0$$
It is correct.
