If $A,B,C$ are $3\times 3$ matrices such that $A,(A-B)$ are invertible and if $(A-B)C=BA^{-1}$, show that $C(A-B)=A^{-1}B$. 
If $A,B,C$ are $3\times 3$ matrices such that $A,(A-B)$ are invertible and if $(A-B)C=BA^{-1}$, show that $C(A-B)=A^{-1}B$.

Usually, $AB$ may not be equal to $BA$. 
I tried starting from the answer.
$$C(A-B)-A^{-1}B+I=I$$
$$C(A-B)-A^{-1}(B-A)=I$$
$$(C+A^{-1})(A-B)=I$$
From the given equation, 
$$(A-B)C-BA^{-1}+I=I$$
$$AC-BC-BA^{-1}+I=I$$
$$A(C+A^{-1})-B(C+A^{-1})=I$$
$$(A-B)(C+A^{-1})=I$$
I can only prove the equation given if I can show that $(C+A^{-1})(A-B)=(A-B)(C+A^{-1})$
How can I do that?

If you have $AB=I$, then it follows $ABA=A$ , and therefore $BA=I$.
So, a matrix always commutes with its inverse. This is exactly what you need to prove your claim.
