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 $a\ne 0$). 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 $a\ne 1$, from $a^3=1$ we get two solutions: $$a=\frac{-1+i\sqrt{3}}{2}\text{and}a=\frac{-1-i\sqrt{3}}{2}.$$ 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.