TITLE: Prove that $\det(AB - BA) = \frac{1}{3}\left(\mathrm{Trace}(AB - BA)^3\right)$
QUESTION [12 upvotes]: I want a correct and feasible answer to this question. 
So does anyone have any creative ideas to prove this equation?

$A$ and $B$ are $3\times3$ matrices.
$\det(AB - BA) = \dfrac{1}{3}\operatorname{Trace}\left((AB - BA)^3\right)$

Answer:
We can write and compute both sides to prove it but this is not a good solution!

REPLY [8 votes]: Another proof.  It is known that for an arbitrary $3 \times 3$ matrix $M$, we have
$$
\det(M) = \frac 16\left[ \operatorname{tr}^3(M) + 2 \operatorname{tr}(M^3)
-3\operatorname{tr}(M)\operatorname{tr}(M^2)\right]
$$
Setting $M = AB - BA$, we note that $\operatorname{tr}(M) = 0$ so that the above simplifies to 
$$
\det(M) = \frac 13 \operatorname{tr}(M^3)
$$
which is the formula that we wanted.