Let $A$ be a Hermitian matrix. Suppose there exists a matrix $B$ such that $A^3B = BA^3$. Show that $AB = BA$.
I was trying to use the fact that since $A$ is Hermitian, there exists a unitary matrix $U$ such that $UDU^* = A$, thus
$UD^3U^* B = BUD^3U^*$,
so $UD^3U^* B - BUD^3U^* = 0$, $D^3U^* BU - U^*BUD^3= 0$ (multiplying both sides by $U^*$ and $U$.
Let $U^* BU = C$, then we have $D^3C = C D^3$, but I get stuck from here...
Suppose the diagonal entries of of $D$ are $\lambda_1,\lambda_2,\ldots, \lambda_n$. Let $p(x)$ be an interpolating polynomial such that $p(\lambda_k^3)=\lambda_k$ for each $k$. Then you can show that $p(A^3)=A$ (e.g. because this holds when restricted to an eigenbasis), and it is clear that $p(A^3)$ commutes with $B$.
This relies on the fact that real cube roots are unique, so that $p$ can be defined as specified. You can see by example that this is false for $A^2$, because $A^2$ might have repeated eigenvalues coming from distinct eigenvalues of $A$. Concretely, this would be the case with $A=\begin{bmatrix}1&0\\0&-1\end{bmatrix}$.
