## Task A-2.1.

Determine all complex numbers $a$ for which all coefficients of the polynomial

$$
P(x)=(x-a)\left(x-a^{2}\right)\left(x-a^{3}\right)
$$

are real.
## First Solution.

Every real number $a$ satisfies the required condition, so let's assume further that $a$ is not a real number. If $z$ is a root of a polynomial with real coefficients, then $\mathrm{i} \bar{z}$ is also a root of that polynomial. Since the polynomial $P(x)$ is of the third degree, it must have at least one real root, so we have only two cases: either $\bar{a}=a^{2}$ or $\bar{a}=a^{3}$.

If $\bar{a}=a^{2}$, then $|a|=|a|^{2}$, and since $a \neq 0$, it follows that $|a|=1$. Furthermore, we get $a^{3}=a^{2} \cdot a=\bar{a} \cdot a=|a|^{2}=1$. For such $a$ we have

$$
\bar{a}=a^{3} \bar{a}=a^{2}|a|^{2}=a^{2}
$$

Thus, the numbers $a$ and $a^{2}$ form a conjugate complex pair, and $a^{3}=1$ is a real number, so we conclude that the polynomial $P(x)$ has real coefficients. From $a^{3}=1$ it follows that $(a-1)\left(a^{2}+a+1\right)=0$, and since $a$ is not a real number, we have $a^{2}+a+1=0$. The solutions to this equation are

$$
a=\frac{-1+i \sqrt{3}}{2} \quad \text { and } \quad a=\frac{-1-i \sqrt{3}}{2}
$$

If $\bar{a}=a^{3}$, similarly we show that $a^{4}=1$. From this, $a= \pm i$, since we are not considering real solutions. By direct verification, we see that both possibilities for $a$ satisfy the conditions of the problem:

$$
P(x)=(x \mp i)(x+1)(x \pm i)=\left(x^{2}+1\right)(x+1)=x^{3}+x^{2}+x+1
$$

Therefore, all complex numbers that satisfy the required condition are all real numbers $a$ and

$$
a \in\left\{i,-i, \frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}\right\}
$$