Solving matrix equation $(X+B)^{-1}=A+BX^{-1}$ for $X$ 
Given that $A,B,X$ are square matrices with same dimensions, find all solutions for $X$ of the equation
  $$(X+B)^{-1}=A+BX^{-1}$$

It is also mentioned that $A,B,X,A+B,B+X,X+A$ are all regular. The first thing I've tried is to group terms which contains $X$ on the LHS and other stuff to the RHS. After multiplying by $X+B$ from the right side I got
$$I=AX+AB+B+BX^{-1}B$$
The problem here is because I have both $X$ and $X^{-1}$ in the linear terms, so I cannot separate it from other terms.
Then I tried to get rid of $X^{-1}$, so I multiplied with $B^{-1}$ from the left
$$B^{-1}-B^{-1}AB-I=B^{-1}AX+X^{-1}B$$
and then by $X$ from the left side
$$X\left(B^{-1}-B^{-1}AB-I\right)=XB^{-1}AX+B$$
How to proceed? Maybe this is a very easy problem which can be solved using some formula similar to quadratic equation, but I am new to matrices, so any help will be appreciated.
 A: Well, yours is not actually an easy problem.
A possible way (probably not the best) to put your equation in a more tractable form could be as follows.
From
$$
\left( {{\bf X + B}} \right)^{\, - \,{\bf 1}}  = {\bf A} + {\bf B}\,{\bf X}^{\, - \,{\bf 1}} 
$$
the LHS can be rewritten as
$$
\begin{array}{l}
 \left( {{\bf X + B}} \right)^{\, - \,{\bf 1}}  = \left( {{\bf X + B}\,{\bf X}^{\, - \,{\bf 1}} {\bf X}} \right)^{\, - \,{\bf 1}}  = \left( {\left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right){\bf X}} \right)^{\, - \,{\bf 1}}  =  \\ 
  = {\bf X}^{\, - \,{\bf 1}} \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right)^{\, - \,{\bf 1}}  \\ 
 \end{array}
$$
which gives:
$$
{\bf X}^{\, - \,{\bf 1}} \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right)^{\, - \,{\bf 1}}  = {\bf A} + {\bf B}\,{\bf X}^{\, - \,{\bf 1}}  = \left( {{\bf A} - {\bf I}} \right) + \left( {{\bf I + B}\,{\bf X}^{\, - \,{\bf 1}} } \right)
$$
Making the substitution
$$
\left\{ \begin{array}{l}
 {\bf Y} = {\bf I + B}\,{\bf X}^{\, - \,{\bf 1}}  \\ 
 {\bf X}^{\, - \,{\bf 1}}  = {\bf B}^{\, - \,{\bf 1}} \left( {{\bf Y} - {\bf I}} \right)\, \\ 
 \end{array} \right.
$$
we arrive at:
$$
{\bf B}^{\, - \,{\bf 1}} \left( {{\bf Y} - {\bf I}} \right) = \left( {{\bf A} - {\bf I}} \right){\bf Y} + {\bf Y}^{\,{\bf 2}} 
$$
i.e.

$$
{\bf Y}^{\,{\bf 2}}  + \left( {{\bf A} - {\bf B}^{\, - \,{\bf 1}}  - {\bf I}} \right){\bf Y} + {\bf B}^{\, - \,{\bf 1}}  = {\bf 0}
$$

