If $$A$$ and $$B$$ are two non-zero $$n \times n$$ matrices such that $$A^2 + B = A^2B$$, then

We are given that $$A$$ and $$B$$ are non-zero $$n \times n$$ matrices satisfying $$A^2 + B = A^2B$$.

Rewriting the given equation, we have $$A^2 + B = A^2B$$, which implies $$A^2 = A^2B - B = (A^2 - I)B$$. It follows that $$A^2(I - B) = -B$$.

If $$(I - B)$$ is invertible, then from $$A^2(I - B) = -B$$ we deduce $$A^2 = -B(I - B)^{-1} = B(B - I)^{-1}$$. By Cayley-Hamilton, $$(B - I)^{-1}$$ is a polynomial in $$B$$ and hence commutes with $$B$$. Therefore,

$$A^2B = B(B - I)^{-1}B = B \cdot B(B - I)^{-1} = B \cdot A^2 = BA^2$$.

The correct answer is Option D: $$A^2B = BA^2$$.