How prove this identity$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$ Today I see a paper,and this author say it is easy to have this identity.But I take sometimes to prove it,and I can't prove it.

show this following identity holds for any real $s$ and $t$ and any postive integer $n$
  $$\sum_{k=0}^{n}\binom{2k}{k}\binom{n+k}{2k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$$

My some idea:
$$\binom{2k}{k}\binom{n+k}{2k}=\dfrac{(2k)!}{(k!)^2}\cdot\dfrac{(n+k)!}{(2k)!(n-k)!}=\dfrac{n!}{k!(n-k)!}\cdot\dfrac{(n+k)!}{n!k!}=\binom{n}{k}\binom{n+k}{k}$$
so we must show this following identity
$$\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(s-t)^{n-k}t^k=\sum_{k=0}^{n}\binom{n}{k}^2s^{n-k}t^k$$
then I can't works
 A: By  way of  enrichment here  is  another algebraic  proof using  basic
complex variables.

For the sum on the LHS which is
$$\sum_{k=0}^n {2k\choose k} {n+k\choose 2k} (s-t)^{n-k} t^k$$
or rather
$$\sum_{k=0}^n  {n\choose k}{n+k\choose k} (s-t)^{n-k} t^k$$
introduce the integral representation
$${n+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+k}}{z^{k+1}} \; dz.$$
This yields for the sum
$$(s-t)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\sum_{k=0}^n {n\choose k}
\left(\frac{1+z}{z} \frac{t}{s-t} \right)^k \; dz
\\ = (s-t)^n \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\left(1+ \frac{1+z}{z} \frac{t}{s-t}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\left(s-t + t \frac{1+z}{z}\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\left(s-t + \frac{t}{z} + t\right)^n \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\left(s + \frac{t}{z}\right)^n \; dz.$$
For the sum on the RHS which is
$$\sum_{k=0}^n {n\choose k}^2 s^{n-k} t^k$$
introduce the integral representation
$${n\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z^{k+1}} \; dz.$$
This yields for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z}
\sum_{k=0}^n {n\choose k} \frac{1}{z^k} s^{n-k} t^k \; dz$$
or
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{z} \left(s+\frac{t}{z}\right)^n \; dz.$$
We have  the identical integrals for  LHS and RHS, done.   We have not
made  use  of  the  properties  of  complex  integrals  here  so  this
computation  can also be  presented using  just algebra  of generating
functions.
There is another computation in the same spirit at this
MSE link.

Apparently  this method is  due to  Egorychev although  some of  it is
probably folklore.
A: $$\begin{align}
&\sum_{k=0}^{n}\color{green}{\binom{2k}k\binom {n+k}{2k}}(s-t)^{n-k}t^k\\
&=\sum_{k=0}^{n}\color{green}{\binom {n+k}{2k}\binom{2k}k}\color{blue}{(s-t)^{n-k}}t^k\\
&=\sum_{k=0}^{n}\color{green}{\binom {n+k}{k}\binom nk} t^k\color{blue}{\sum_{r=0}^{n-k}\binom{n-k}rs^{n-k-r}(-t)^r}\\ 
&=\color{orange}{\sum_{k=0}^{n}\sum_{r=0}^{n-k}}\binom{n+k}k\binom nk\binom{n-k}rs^{n-k-r}t^{k+r}(-1)^r\\
&=\color{orange}{\sum_{j=0}^{n}\sum_{k=0}^{j}}\binom{n+k}k\color{purple}{\binom nk\binom{n-k}{j-k}}s^{n-j}t^{j}(-1)^{j-k}\\
&=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{n+k}k}\color{purple}{\binom n{n-k}\binom{n-k}{n-j}}(-1)^{j-k}\\
&=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\color{red}{\binom{-n-1}k (-1)^k}\color{purple}{\binom n{n-j}\binom{j}{k}}(-1)^{j-k}\\
&=\sum_{j=0}^{n}s^{n-j}t^j\sum_{k=0}^{j}\binom{-n-1}k\color{purple}{\binom nj\binom j{j-k}}(-1)^{j}\\
&=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\sum_{k=0}^{j}\binom{-n-1}k\binom j{j-k}}\\
&=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom{-n-1+j}j}\\
&=\sum_{j=0}^{n}\binom nj(-1)^{j}s^{n-j}t^j\color{darkgreen}{\binom nj (-1)^j}\\
&=\sum_{j=0}^{n}\color{darkred}{\binom nj}s^{n-j}t^j\color{darkred}{\binom nj}\\
&=\sum_{j=0}^{n}\color{darkred}{{\binom nj}^2} s^{n-j}t^j\\
&=\sum_{k=0}^{n}{\binom nk}^2 s^{n-k}t^k\qquad \blacksquare 
\end{align}$$

NB: The above also shows that
$$\small\begin{align}
&\sum_{k=0}^{j}\underbrace{\binom{n+k}k\binom nk\binom{n-k}{j-k}\color{gray}{\binom{n-j}{n-j}}}_{\Large\binom{n+k}{k,k,j-k,n-j}}(-1)^{j-k}\\
&=\sum_{k=0}^{j}\binom{n+k}{k,k,j-k,n-j}(-1)^{j-k}\\
&={\binom nj}^2
\end{align}$$
