$A, B$ are $3 \times 3$ matrices such that $(A - B)^2 = 0$. Prove that $\operatorname{Tr}(AB - BA)^3 = 0$. I have been trying to solve this recent linear algebra problem:

Let $A, B$ be $3 \times 3$ matrices such that $(A-B)^2 = 0$. Prove that $\det (AB - BA) = 0$.

This was my approach:$\DeclareMathOperator{\Tr}{Tr}$
The following equality holds for any $3\times 3$ matrices $A, B$:
$$\det (AB - BA) = \frac13 \Tr(AB - BA)^3$$
It follows from the Hamilton-Cayley theorem applied on $AB - BA$. Therefore, it suffices to prove that $\Tr(AB - BA)^3 = 0$.
Expanding gives that $\Tr(AB - BA)^3$ is equal to:
\begin{align}
\Tr\left(\color{magenta}{ABABAB} - \color{blue}{ABABBA} - \color{purple}{ABBAAB} + \color{purple}{ABBABA} - \color{blue}{BAABAB} + \color{olive}{BAABBA} + \color{olive}{BABAAB} - \color{magenta}{BABABA}\right)\\
\end{align}
where the same-colored terms are cyclic permutations of each other so have the same trace.
So, $$\Tr(AB - BA)^3 = 2\Tr BAABBA - 2\Tr BAABAB = 2 \Tr BAAB(BA - AB)$$
I figured this was a good place to try to use the assumption $(A - B)^2 = 0$:
$$0 = (A - B)^2 = A^2 + B^2 - AB - BA \implies A^2 = AB + BA - B^2$$
So we have:
$$BAAB(BA - AB) = B(AB + BA - B^2)B(AB - BA) = $$
$$\color{OrangeRed}{BABBBA} + \color{green}{BBABBA} - BBBBBA - \color{green}{BABBAB} - \color{OrangeRed}{BBABAB} + BBBBAB$$
Again, the same-colored terms cancel out when taking the trace so:
$$\Tr BAAB(BA - AB) = \Tr BBBB(AB - BA)$$
A possible development is:
\begin{align}2 \Tr BAAB(BA - AB) &=  \Tr (AB-BA)(BBBB - BAAB)\\
&= \Tr (AB-BA)(BBBB - BAAB)\\
&= \Tr (AB-BA)B(B^2 - A^2)B
\end{align}
But using $A^2 + B^2 = AB + BA$ here again gives $\Tr (BA - AB)BAAB$, so nothing new.
Is there a way to finish the proof?
 A: Let $N=A-B$. Then $N^2=0$ and $C=AB-BA=NB-BN$. Therefore
\begin{align}
C^2&=(NB-BN)(NB-BN)=NBNB-NBBN+BNBN,\\
C^3&=(NB-BN)(NBNB-NBBN+BNBN)\\
&=NBNBNB-NBNBBN+NBBNBN-BNBNBN.\tag{1}
\end{align}
Now, using the cyclic property of matrix trace $\operatorname{tr}(XY)=\operatorname{tr}(YX)$ and the assumption that $N^2=0$, we immediately see that the two middle summands on line $(1)$ are traceless and the traces of the two outer summands cancel out each other. Hence $\operatorname{tr}(C^3)=0$.
A: Forget $A$ and $B$ are $3\times 3$ matrices. Let's say $A, B$ are $n \times n$
matrices satisfying the algebraic relation $(A-B)^2 = 0$. Let $X = A-B$, then $X^2 = 0$ and
$$AB-BA = (B+X)B - B(X+B) = XB-BX$$
Let $P = XB = (A-B)B$ and $Q = BX = B(A-B)$, we have $$QP = BXXB = B(A-B)^2B = 0$$
For any integer $m > 1$, when we expand $(AB-BA)^m = (P-Q)^m$ into sum of monomials in $P,Q$. If $Q$ appear before any $P$, then that term vanishes. As a result, only terms that all $P$ is on the left of $Q$ survives. More precisely,
$$(AB-BA)^m = \sum_{\ell=0}^m (-1)^\ell P^{m-\ell} Q^\ell = P^m - P^{m-1}Q + P^{m-2}Q^2 + \cdots + (-1)^m Q^m$$
For terms that contains both $P$ and $Q$ (i.e $0 < \ell < m$), we have
$${\rm Tr}(P^{m-\ell}Q^\ell) = {\rm Tr}(Q^\ell P^{m-\ell})
= {\rm Tr}(Q^{\ell-1}(QP)P^{m-\ell-1}) = {\rm Tr}(Q^{\ell-1} 0_n P^{m-\ell-1}) = {\rm Tr}(0_n) = 0$$
This leads to
$${\rm Tr}(AB-BA)^m = {\rm Tr}(P^m + (-1)^mQ^m)$$
When $m$ is odd, this becomes
$${\rm Tr}((AB-BA)^m) = {\rm Tr}((XB)^m - (BX)^m)
= {\rm Tr}(X(BX)^{m-1}B - (BX)^{m-1}BX)\\
= {\rm Tr}([X,(BX)^{m-1}B]) = 0$$
When $m = 3$, this reduces to the identity you wish to show.
