Let
$X=AB-BA.$
We know that for any square matrices $A$ and $B$,
$\operatorname{tr}(AB)=\operatorname{tr}(BA).$
Therefore,
$\operatorname{tr}(X)=\operatorname{tr}(AB-BA)=\operatorname{tr}(AB)-\operatorname{tr}(BA)=0.$
Now let the eigenvalues of $X$ be $\lambda_1,\lambda_2,\lambda_3$.
Since $X$ is a $3 \times 3$ matrix, we have
$\lambda_1+\lambda_2+\lambda_3=\operatorname{tr}(X)=0.$
Also,
$\det(X)=\lambda_1\lambda_2\lambda_3$
and
$\operatorname{tr}(X^3)=\lambda_1^3+\lambda_2^3+\lambda_3^3.$
Using the identity
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca),$
and putting $a=\lambda_1$, $b=\lambda_2$, $c=\lambda_3$, we get
$\lambda_1^3+\lambda_2^3+\lambda_3^3=3\lambda_1\lambda_2\lambda_3$
because
$\lambda_1+\lambda_2+\lambda_3=0.$
Hence,
$\operatorname{tr}(X^3)=3\det(X).$
So,
$\det(X)=\frac{\operatorname{tr}(X^3)}{3}.$
Since $X=AB-BA$, this gives
$\det(AB-BA)=\dfrac{\operatorname{tr}\left((AB-BA)^3\right)}{3}.$
Therefore the given statement is $\textbf{True}.$
