Let  A and B be $3\times 3$  matrices, then  $\frac{tr\left[{(AB-BA)}^3\right]}{\det (AB-BA)}=$

Let  $AB-BA=X_{3\times 3}$  Use Cayley – Hamilton theorem. 
$X^3-C_1{X}^2+C_{2}X-C_3{I}_3=O_{3}where{C}_1=trX=tr(AB-BA)=0$  
$C_3=\det \left(X\right),takingtraceoftheaboveequation,wehave,$ 
$tr\left(X^3\right)-3\det \left(X\right)=0\Rightarrow \frac{tr\left(X^3\right)}{\det \left(X\right)}=3$