$\text{Both matrices $A$ and $B$ belong to the set of real matrices of size $3\times 3$}$
$\text{Suppose, $AB-BA = C$ and eigen values of C are $\lambda_1$,$\lambda_2$ and $\lambda_3$}$
$\text{Trace($(AB-BA)^3$)= Trace($C^3)$}$
$= \lambda_1^3 +\lambda_2^3 + \lambda_3^3 = (\lambda_1+\lambda_2 +\lambda_3)(\lambda_1^2+\lambda_2^2 + \lambda_3^2 – \lambda_1\lambda_2-\lambda_2\lambda_3-\lambda_3\lambda_1) + 3 \lambda_1 \lambda_2 \lambda_3$
$\text{Here, $\lambda_1+\lambda_2 +\lambda_3$ = Trace$(AB-BA)=$ Trace$(AB)-$Trace$(BA)=0$}$ $(\because Trace(AB) = Trace(BA) )$
$\text{So, Trace($C^3)= 3 \lambda_1 \lambda_2 \lambda_3 = 3 $ det$(C)$}$
$\text{Hence, $\frac{Trace((AB-BA)^3)}{3} = det(AB-BA)$}$