Problem:
I want to prove that if $A^3$ is unitary, then $A$ is diagonalizable. Definitely, since $A^3$ is unitary, then it is normal, and we know that every normal matrix is diagonalizable. So, there exists $U$ unitary such that: $U^{*}A^{3}U=D=\begin{bmatrix} \lambda_{1}^{3}&0 &0 &0 \\ 0&\lambda_{2}^{3} &0 &0 \\ 0 & 0 & \lambda_{3}^{3}&0 \\ 0& 0 &0 & \lambda_{n}^{3} \end{bmatrix}$
Where $\lambda_{i}$ are eigenvalues of $A$. I am thinking to prove that $A$ is also normal, which implies that $A$ is diagonalizable, but I have no idea how to approach it.