## Task A-2.1.

Let $a$ be a complex number such that

$$
a^{5}+a+1=0
$$

What values can the expression $a^{2}(a-1)$ take?
## Solution.

It holds that

$$
\begin{aligned}
a^{5}+a+1 & =\left(a^{5}-a^{4}\right)+\left(a^{4}-a^{3}\right)+\left(a^{3}-a^{2}\right)+a^{2}+a+1 \\
& =\left(a^{2}+a+1\right)\left(a^{3}-a^{2}+1\right)
\end{aligned}
$$

If $a^{2}+a+1=0$, then $a^{2}(a-1)=-(a+1)(a-1)=1-a^{2}=a+2$.

Since in this case $a=\frac{-1 \pm i \sqrt{3}}{2}$, we have $a^{2}(a-1)=\frac{3}{2} \pm \frac{\sqrt{3}}{2} i$.

If $a^{3}-a^{2}+1=0$, then $a^{2}(a-1)=a^{3}-a^{2}=-1$.

The expression under consideration can take the values $-1, \frac{3}{2}+\frac{\sqrt{3}}{2} i \text{ and } \frac{3}{2}-\frac{\sqrt{3}}{2} i$.