Given,

$A+B=\left[\begin{array}{lll}2& 1& 2\\ 1& 2& 0\\ 0& 2& 2\end{array}\right] \& AB=\left[\begin{array}{lll}1& 2& 2\\ 1& 1& 0\\ 1& 2& 1\end{array}\right]$

Now we know that 

${\left(A+B\right)}^2=A^2+AB+BA+B^2$

$\Rightarrow {\left(A+B\right)}^2-AB=A^2+BA+B^2$

$\Rightarrow {\left(A+B\right)}^2-AB=A^2+B\left(A+B\right)$

So $A^2+B\left(A+B\right)={\left(A+B\right)}^2-AB ...\left(1\right)$

Now finding ${\left(A+B\right)}^2=\left[\begin{array}{lll}2& 1& 2\\ 1& 2& 0\\ 0& 2& 2\end{array}\right]\times \left[\begin{array}{lll}2& 1& 2\\ 1& 2& 0\\ 0& 2& 2\end{array}\right]$

$\Rightarrow {\left(A+B\right)}^2=\left[\begin{array}{lll}5& 8& 8\\ 4& 5& 2\\ 2& 8& 4\end{array}\right]$

Now putting the value of ${\left(A+B\right)}^2 \& AB$ in equation $\left(1\right)$ we get,

$A^2+B\left(A+B\right)=\left[\begin{array}{lll}5& 8& 8\\ 4& 5& 2\\ 2& 8& 4\end{array}\right]-\left[\begin{array}{lll}1& 2& 2\\ 1& 1& 0\\ 1& 2& 1\end{array}\right]=\left[\begin{array}{lll}4& 6& 6\\ 3& 4& 2\\ 1& 6& 3\end{array}\right]$