We are tasked with solving for the matrix $X$ in the equation:

$A X (D + B X)^{-1} = C$

Assume all matrices are $n \times n$ and invertible where necessary.

Step-by-Step Solution:

1. Multiply both sides by $(D + BX)$ from the right: To eliminate the inverse, multiply both sides of the equation on the right by $D + BX$: $A X (D + B X)^{-1}(D + BX) = C (D + BX)$ Simplifying the left-hand side: $A X = C (D + BX)$

2. Expand the right-hand side: Distribute $C$ over $(D + BX)$: $A X = C D + C B X$

3. Group the terms with $X$: Move the term with $X$ on the right-hand side to the left-hand side: $A X - C B X = C D$ Factor out $X$ on the left-hand side: $(A - C B) X = C D$

4. Solve for $X$: Assuming $A - C B$ is invertible, multiply both sides by $(A - C B)^{-1}$ from the left: $X = (A - C B)^{-1} C D$

Final Expression for $X$:

$X = (A - C B)^{-1} C D$

Would you like further details or clarification on any step? Here are some related questions:

1. How do we determine if $A - CB$ is invertible?
2. What if $A - CB$ is not invertible? How would the solution change?
3. Can this method be generalized to other matrix equations involving inverses?
4. How would the process differ if $B$ or $D$ were non-square matrices?
5. What real-world applications might involve solving matrix equations like this?

Tip: When working with matrix inverses, always ensure that the matrix in question is invertible by checking its determinant or other criteria for non-singularity.