Proof with invertible matrices Let $A,B,C$ be matrix of the same size, and suppose A is invertible, Prove that
$(A-B)C=BA^{-1}$ then $C(A-B)=A^{-1}B$
I tried to prove it as following.
$(A-B)C=BA^{-1}$ implies
$AC-BC=BA^{-1}$ so
$ACA-BCA=B$ taking $A^{-1}$
$CA-A^{-1}BCA=A^{-1}B$
Any hint will be appreciated

Assume that $(A-B)C = BA^{-1}$. Then $A-B$ is invertible, and its inverse is $C+A^{-1}$. You can check:
$$ (A-B)(A^{-1}+C) = \mathrm{Id} + AC -BA^{-1} -BC = \mathrm{Id} + \bigg( (A-B)C - BA^{-1} \bigg) $$
But the assumption implies that the part in parentheses is zero, so this product is just the identity matrix. Inverse matrices always commute. By that I mean $X X^{-1} = X^{-1}X$ whenever $X$ is invertible. So you can re-arrange the product above in the opposite order:
$$ (A^{-1} + C)(A-B) = \mathrm{Id} $$
If you expand this out and re-arrange the terms, you'll get the expression you want:
$$ C(A-B) = A^{-1}B $$
How I came up with this solution:
The first thing I thought to do was a "change of variables". I thought let's give a name to $A-B$ and treat it as one single matrix. So I called it $X$:
$$ X := A-B $$
So then I re-wrote the original equation $(A-B)C = BA^{-1}$ using $X$ instead of $B$:
$$ XC = (A-X)A^{-1} = \mathrm{Id} - XA^{-1} $$
If you move the negative term to the other side and factor, you get
$$ X(C+A^{-1}) = \mathrm{Id} $$
