In the exercise I have to find the exact value of $\sum_{k=0}^{n}(-1)^k\binom{n}{k}^2$. Now I already have seen find $\sum_{k=0}^{t}(-1)^k\binom{t}{k}^2$ for odd t then for even t
and I do not understand the first solution. That's why I am writing. I am kind of confused from where does this guy get this: $\sum_{k=0}^{t}\binom{t}{k}^2(-1)^k = [x^t]\left[\left(\sum_{k=0}^{t}\binom{t}{k}(-1)^k x^k\right)\cdot\left(\sum_{k=0}^{t}\binom{t}{t-k}x^{t-k}\right)\right]$
All I figured out myself is that I have to somehow use the fact that $(1-x^2)^n=(1-x)^n(1+x)^n$. And also all I did was to expand those two sides of an equation using binomial theorem. Now I am lost and do not know how to proceed further. Can somebody please give me a clear explanation what should I do next?
Thanks in advance
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}^{2} & = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{n \choose n - k} = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{z^{n - k}}\pars{1 + z}^{n} \\[5mm] & = \bracks{z^{n}}\pars{1 + z}^{n}\sum_{k = 0}^{n}{n \choose k}\pars{-z}^{k} = \bracks{z^{n}}\pars{1 + z}^{n}\pars{1 - z}^{n} = \bracks{z^{n}}\pars{1 - z^{2}}^{n} \\[5mm] & = \bracks{z^{n}}\sum_{k = 0}^{n}{n \choose k}\pars{-z^{2}}^{k} = \bbx{\begin{array}{lcl} \ds{{n \choose n/2}\pars{-1}^{n/2}} & \mbox{if} & \ds{n}\ \mbox{is}\ even \\[2mm] \ds{0} && otherwise \end{array}} \end{align}