In general, matrix multiplication is not commutative: $AB$ and $BA$ might be different.

Solution.

Let us calculate $(A-B)(A+B)$ as follows using the fact that the matrix product is distributive.
\begin{align*}
(A-B)(A+B)&=A(A+B)-B(A+B)\\
&=A^2+AB-BA-B^2\\
&=A^2-B^2+(AB-BA).
\end{align*}
Thus if $(A-B)(A+B)=A^2-B^2$ then $AB-BA=O$, the zero matrix. Equivalently, $AB=BA$.

Note that matrix multiplication is not commutative, namely, $AB\neq BA$ in general.
Thus we can disprove the statement if we find matrices $A$ and $B$ such that $AB \neq BA$.

For example, let
\[A=\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
\text{ and }
B=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}.\]
Then we have
\[AB=\begin{bmatrix}
0 & 2\\
0& 0
\end{bmatrix} \text{ and }
BA=\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}.\]

Since $AB \neq BA$, we have $(A-B)(A+B) \neq A^2-B^2$ for \[A=\begin{bmatrix}
1 & 1\\
0& 0
\end{bmatrix}
\text{ and }
B=\begin{bmatrix}
0 & 1\\
0& 1
\end{bmatrix}.\]

Hence we found a counterexample for the statement.

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