On summation $\sum_{k = 0}^n \binom nk^2 x^k y^{n-k} $

Question

I already know that $$\sum_{k = 0}^n \binom nk^2=\binom{2n}{n}.$$ But, do we know anything about the summation $$\sum_{k = 0}^n \binom nk^2 x^k y^{n-k} $$ for arbitrary $x$ and $y$? Thanks.

Answer 1

Note the identity:

$$S_n(x,y)=y^n\sum_{k=0}^{n} {n \choose k}^2 (x/y)^k= y^n \left(1-\frac{x}{y}\right)^n P_n\left(\frac{y+x}{y-x}\right),$$ where $P_n(z)$ are Legendre polynomials of order $n$.