How to implement the formula without computing any matrix inverse? Given $A, B, C$ are $n \times n$ matrices, $B, C$ are nonsingular and $b$ is an $n$-vector, and $$x=B^{-1}(2A+I)(C^{-1}+A)b$$ This is a question coming from my computational mathematics course, and right now we are learning LU factorization. However, I don't see it can be used to solve this question. I also tried to move the elements to the other side, but things were getting too complicated to solve. Any answers will be appreciated!

I actually got the answer myself. Since $$x=B^{-1}(2A+I)(C^{-1}+A)b$$ By multiplying $B$ on both sides and opening the brackets, we get $$Bx=2AC^{-1}b+2AAb+C^{-1}b+Ab$$ Let $k$ be an n-vector, $k=C^{-1}b$, then $Ck=b$. By LU factorization, we can get $k$. Plug $k$ into the equation above, we get $$Bx=2Ak+2AAb+k+Ab$$ Let $y=2Ak+2AAb+k+Ab$, $y$ is an n-vector. Thus we get$$Bx=y$$Again, by LU factorization, we can find out $x$. QED.
