$\text{ The constant term in the expansion of }{\left(x^2+\frac{1}{x^2}+y+\frac{1}{y}\right)}^8\text{ is }$

$\begin{array}{l}\text{ Given expansion }{\left(x^2+\frac{1}{x^2}+y+\frac{1}{y}\right)}^8\\ =\sum \frac{8!}{\alpha !\beta !\gamma !\delta !}{x}^{2\alpha }\frac{1}{x^{2\beta }}{y}^{\gamma }\frac{1}{y^{\delta }}\\ \text{ Here }\alpha +\beta +\gamma +\delta =8\\ \alpha =\beta ,\gamma =\delta \\ \alpha =0,\beta =0\gamma =4,\delta =4\\ \alpha =1,\beta =1\gamma =3,\delta =3\\ \alpha =2,\beta =2\gamma =2,\delta =2\\ \alpha =3,\beta =3\gamma =1,\delta =1\\ \alpha =4,\beta =4\gamma =0,\delta =0\\ \text{ Constant term in the expression }=2\left(\frac{8!}{0!0!4!4!}\right)+2\left(\frac{8!}{1!1!3!3!}\right)+\frac{8!}{2!2!2!2!}=4200\end{array}$