Prove $BA - A^2B^2 = I_n$. I have a problem with this. Actually, still don't have the right way to start :/
Problem :
Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$.
Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.
Thanks for any help.

First, I try to demonstrate that $A$ and $B$ are invertible.
Let $(\lambda_j, v_j)$ be a couple of eigenvalue-eigenvector of $A$. Then, $(\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $A^3$. Since $A^3 = -B^3$, then we can say that $(-\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $B^3$. Finally, we can state that $(-\lambda_j, v_j)$ is is a couple of eigenvalue-eigenvector of $B$.
Consider now the following equations:
$$ABv_j - B^2A^2v_j = I_n v_j$$
$$A (-\lambda_j)v_j - B^2 (\lambda_j^2)v_j = v_j$$
$$(\lambda)_j (-\lambda_i)v_j - (\lambda_j^2) (\lambda_j^2)v_j = v_j$$
$$(\lambda_j^4 + \lambda_j^2 + 1)v_j = 0$$
Posing $(\lambda_j^4 + \lambda_j^2 + 1) = 0$, we get that eigenvalues are $\frac{1 \pm i \sqrt{3}}{2}$ and $\frac{-1 \pm i \sqrt{3}}{2}$. This means that neither $A$ nor $B$ has a null eigenvalues and hence $A$ and $B$ are both invertible.
At this point, we have that:
$AB - B^2A^2 = I_n \Rightarrow A = (I_n + B^2A^2)B^{-1}$
Then:
$BA - A^2B^2 = B(I_n + B^2A^2)B^{-1} - A^2B^2 = I_n + B^3A^2B^{-1} -A^2B^2$
We know that $A^3 = -B^3$, and then:
$BA - A^2B^2 = I_n - A^3A^2B^{-1} -A^2B^2 = I_n + A^2 B^3 B^{-1} - A^2B^2 = I_n + A^2B^2 - A^2B^2 = I_n$.
