Prove that $\det(AB - BA) = \frac{1}{3}\left(\mathrm{Trace}(AB - BA)^3\right)$ 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!

This follows easily from Cayley-Hamilton theorem. Since $M=AB-BA$ has zero trace, by Cayley-Hamilton theorem, $M^3=cM+dI_3$ where $c$ is some scalar and $d=\det(M)$. Therefore $\operatorname{tr}(M^3)=3d$ and the result follows.
