matrix equation $(A-B)CA=B$ let $A,B,C$ be $n\times n$ matrices with real entries such that $A$ is invertible. if $(A-B)CA=B$ show that $AC(A-B)=B$.
any Ideas??

Let $Q = A-B$.  Since $QCA=B = A-Q$, $Q(CA+I)=A$, and therefore $Q$ is invertible.
Now $C = Q^{-1} B A^{-1} = Q^{-1}(A-Q)A^{-1} = Q^{-1} - A^{-1}$, so 
$AC(A-B) = A(Q^{-1} - A^{-1}) Q = A - Q = B$.
