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.

By the two given conditions, we see that
\begin{align*}
A(BA-A^2B^2) &= ABA-A^3B^2\\
&=(I+B^2A^2)A-A^3B^2\\
&=A+B^2A^3-A^3B^2\\
&=A-B^5+B^5\\
&=A.
\end{align*}
Therefore, if we can prove that $A$ is invertible, we are done.
Suppose the contrary. Then there exists a nonzero vector $v$ such that $Av=0$. We now prove by mathematical induction that for $k\ge1$, $AB^kv = a_kB^{k-1}v$ for some $a_k\neq0$. The base case is easy: $ABv=(I+B^2A^2)v=v$. Now, for $k\ge1$,
\begin{align*}
AB^{k+1}v
&= (AB)B^kv\\
&= (B^2A^2+I)B^kv\\
&= B^2A(AB^kv) + B^kv\\
&= B^2A(a_kB^{k-1}v) + B^kv\\
&= a_kB^2(AB^{k-1}v) + B^kv\\
&= \begin{cases}
0+B^kv &\text{ when } k=1\\
a_k B^2 (a_{k-1}B^{k-2}v) + B^kv &\text{ when } k\ge2
\end{cases}\\
&= \begin{cases}
B^kv &\text{ when } k=1\\
(a_{k-1}a_k+1) B^kv &\text{ when } k\ge2
\end{cases}.
\end{align*}
In short, if we define $a_0=0,\ a_1=1$ and $a_k=a_{k-2}a_{k-1}+1$ for $k\ge2$, we have
\begin{align*}
Av &= 0,\tag{1}\\
AB^kv &= a_kB^{k-1}v\ \text{ for }\ k\ge1\tag{2}
\end{align*}
with $a_k\neq0$ for $k\ge1$.
Let $m\ge0$ be the largest integer such that $v,Bv,B^2v,\ldots,B^mv$ are linearly independent, and let $B^{m+1}v=\sum_{k=0}^m c_kB^kv$. By $(1)$ and $(2)$,
$$
a_{m+1}B^mv = AB^{m+1}v
= A \sum_{k=0}^m c_k B^k v
= \sum_{k=\color{red}{1}}^m c_k AB^k v
= \sum_{k=1}^m c_ka_k B^{k-1} v.\tag{3}
$$
Yet this is impossible because $v,Bv,B^2v,\ldots,B^mv$ are linearly independent and $a_{m+1}\neq0$. Hence the assumption that $v$ is nonzero cannot be true and $A$ is invertible.
