Croatian Mathematical Society Competitions

Obviously, all real numbers satisfy the given condition.
Let us, from now on, assume that $a$ is not a real number.
Since $P$ is a polynomial of degree 3, it must have at least one real root. We also know that if $z$ is a root of $P$, then $\bar{z}$ is a root of $P$ as well. Hence $\bar{a} = a^2$ or $\bar{a} = a^3$.
In the first case, we get $|a| = |a|^2$, and then $|a| = 1$ (since $a \neq 0$). Hence $a^3 = a^2 \cdot a = \bar{a} \cdot a = |a|^2 = 1$, so both factors $(x - a)(x - a^2)$ and $x - a^3$ of $P$ have real coefficients. Since $a \neq 1$, from $a^3 = 1$ we get two solutions:
$$a = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad a = \frac{-1 - i\sqrt{3}}{2}.$$
In the second case, we similarly get $a^4 = 1$, hence $a = \pm i$, so both factors $(x-a)(x-a^3)$ and $x-a^2$ of $P$ have real coefficients.

$$a \in \left\{ i, -i, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2} \right\},$$
along with all real numbers.