Answer
$AB\neq BA$.
Work Step by Step
We have
\begin{align*}
(A+B)(A+B)&=A^2+AB+BA+B^2 \\
&\neq A^2+2AB++B^2 .
\end{align*}
This because, $AB\neq BA$ in general. For example, consider
$$A=\left[\begin{array}{cc}{1} &{2} \\{0}& {1} \end{array}\right], \quad B=\left[\begin{array}{cc}{1} &{0} \\{-1}& {1} \end{array}\right].$$
Now, one can see that
$$AB=\left[\begin{array}{cc}{1} &{2} \\{0}& {1} \end{array}\right]\left[\begin{array}{cc}{1} &{0} \\{-1}& {1} \end{array}\right]=\left[\begin{array}{cc}{-1} &{2} \\{-1}& {1} \end{array}\right]$$
$$BA=\left[\begin{array}{cc}{1} &{0} \\{-1}& {1} \end{array}\right]\left[\begin{array}{cc}{1} &{2} \\{0}& {1} \end{array}\right]=\left[\begin{array}{cc}{1} &{2} \\{-1}& {-1} \end{array}\right]$$
Hence, $AB\neq BA$.
