Solution:

Output due to weighted sum of inputs:

$y_3(n)=a x_1(n-2)+b x_2(n-2)+a x_1\left(n^2\right)+b x_2\left(n^2\right)\\ $

$\begin{aligned} & y_1(n)=x_1(n-2)+x_1\left(n^2\right) \\\\ & y_2(n)=x_2(n-2)+x_2\left(n^2\right) \\ \end{aligned}$

$\begin{aligned}\\ a y_1(n)+b y_2(n)= & a x_1(n-2)+a x_1\left(n^2\right)+b x_2(n-2)+b x_2\left(n^2\right) \\\\ & \because y_3(n)=a y_1(n)+b y_2(n)\\ \end{aligned}$