How to show that $A(A+B)^{-1}B = B(A+B)^{-1}A$ I am trying this problem. Could not proceed . Please give a hint.
The exercise, if the above image is not showing clearly, is the following: Let $A$ and $B$ be two square $m \times m$ matrices for some integer $m \ge 2$ such that $A+B$ is invertible. Then show that
the equation $A(A+B)^{-1}B = B(A+B)^{-1}A$. [Note that either $A$ or $B$ or each of both may not be invertible, and that $A$ and $B$ may not commute.]
First note the following equation:
$$A(A+B)^{-1}(A+B) = (A+B)(A+B)^{-1}A.$$
[Indeed, both sides of this equation just above are $A$.] That hint may or may not be enough. If you want more help, see below:

 Distributing on each side gives.....


 $$A(A+B)^{-1}A +A(A+B)^{-1}B = A(A+B)^{-1}A + B(A+B)^{-1}A.$$


 But then subtracting both sides by $A(A+B)^{-1}A$ gives


 $$A(A+B)^{-1}B = B(A+B)^{-1}A,$$


 which is what you want.
You can try prove:
$$
A(A+B)^{-1} B * (B(A+B)^{-1}A)^{-1}=I
$$