<s>[INST] Complex numbers, polynomials Let $a$ be complex number such that $a^5 + a + 1 = 0$. What are possible values of $a^2(a - 1)$?    
I have tried to find $a$. Is there any way to find it?
 [/INST] $$
\begin{align}
a^5+a+1&=(a^2+a+1)(a^3-a^2+1)\\
&=(a^2+a+1)(a^2(a-1)+1)
\end{align}$$
If $a$ vanishes the second factor then $a^2(a-1)=-1$.
If $a$ vanishes the first factor, then divide $a^2(a-1)$ by $a^2+a+1$. We get $$a^3-a^2 = (a-2) × (a^2+a+1)+a+2$$
Therefore $a^2(a-1)=a+2$. We know that the roots of $a^2+a+1$ are $\frac{-1\pm\sqrt{1-4}}{2}$. Then only need to add $2$ to them.
In conclusion, the possible values are 
$$-1, \frac{3\pm\sqrt{-3}}{2}.$$
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