<s>[INST] How to show that $\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$ How to show that 

$$\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$$

 [/INST] Here is an answer  based upon  Short Proofs of Saalschütz's and Dixon's theorems by Ira Gessel and Dennis Stanton.

The answer is provided in three steps
  
  
*
  
*Step 1: Relationship between coefficients of the constant term of a bivariate Laurent series and a transformed of it.
  
*Step 2: Application of this relationship to a specific Laurent series with the coefficient of the constant term being strongly related with OPs binomial identity.
  
*Step 3: Substitutions to reveal the symmetry of the binomial identity and to transform it finally to OPs stated binomial identity. 

We consider formal Laurent series
\begin{align*}
A(x,y)=\sum_{m,l}a_{m,l}x^my^l
\end{align*}
in which for only finitely many pairs $(m,l)$ the coefficients $a_{m,l}\ne 0$ with $m<0$ or $l<0$. Let $[x^n]$ denote the coefficient of $x^n$ in a series. We start with a useful observation

Step 1: Let $A(x,y)$ be a formal Laurent series. The following is valid
  \begin{align*}
[x^0y^0]A\left(\frac{x}{1+y},\frac{y}{1+x}\right)=[x^0y^0]\frac{1}{1-xy}A(x,y)\tag{1}
\end{align*}
By linearity it is sufficient to prove (1) by considering $A(x,y)=x^ly^m$ with $l,m\in \mathbb{Z}$. We have to show
  \begin{align*}
[x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}=[x^0y^0]\frac{x^ly^m}{1-xy}\tag{2}
\end{align*}

We show the LHS of (2) fulfills 
\begin{align*}
[x^0y^0]\frac{x^l}{(1+y)^l}\cdot\frac{y^m}{(1+x)^m}=
\begin{cases}
1&\qquad\text{if }l=m\leq 0\\
0&\qquad\text{otherwise}
\end{cases}\tag{3}
\end{align*}


*

*If $l>0$ or $m>0$ the result in (3) is $0$ since the binomial expansion of $(1+y)^{-l}(1+x)^{-m}$ has no negative powers.

*If $l=-r,m=-s$ with $r,s\geq 0$ we note that
\begin{align*}
[x^0y^0]\frac{(1+x)^s(1+y)^r}{x^ry^s}&=[x^ry^s](1+x)^s(1+y)^r\\
&=\binom{s}{r}\binom{r}{s}=
\begin{cases}
1&\qquad\text{if }r=s\\
0&\qquad\text{if }r\neq s
\end{cases}
\end{align*}
and (3) follows.
We look at the RHS of (2) and observe
\begin{align*}
[x^0y^0]\frac{x^ly^m}{1-xy}=[x^{-l}y^{-m}]\sum_{j\geq 0}(xy)^j
=\begin{cases}
1&\qquad\text{if }l=m\leq 0\\
0&\qquad\text{otherwise}
\end{cases}
\end{align*}
which is the same as (3) and the claim (2) follows.

Step 2: We now apply (1) to the Laurent series
  \begin{align*}
f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^n}\qquad\qquad\qquad l,m\in\mathbb{Z},n\in\mathbb{N}
\end{align*}
We obtain
\begin{align*}
&f\left(\frac{x}{1+y},\frac{y}{1+x}\right)\\
&\qquad=\left(\frac{x}{1+y}-\frac{y}{1+x}\right)^n
\cdot\frac{(1+y)^l}{x^l}\cdot\frac{(1+x)^m}{y^m}\cdot\left(1-\frac{xy}{(1+x)(1+y)}\right)^{-n}\\
&\qquad=\frac{\left((x-y)+(x^2-y^2)\right)^n}{(1+x)^n(1+y)^n}\cdot
\frac{(1+y)^l}{x^l}\frac{(1+x)^m}{y^m}\cdot
\frac{(1+x)^n(1+y)^n}{(1+x+y)^n}\\
&\qquad=\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\tag{4}
\end{align*}
on the other hand we get
\begin{align*}
\frac{1}{1-xy}f(x,y)=\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{5}
\end{align*}
equating (4) and (5) we obtain according to (2)
  \begin{align*}
[x^0y^0]\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}=[x^0y^0]\frac{(x-y)^n}{x^ly^m(1-xy)^{n+1}}\tag{6}
\end{align*}

The LHS of (6) gives
\begin{align*}
[x^0y^0]&\frac{(1+x)^m(1+y)^l(x-y)^n}{x^ly^m}\\
&=[x^ly^m]\sum_{k}\binom{n}{k}(-y)^kx^{n-k}(1+x)^m(1+y)^l\\
&=\sum_{k}\binom{n}{k}(-1)^k[x^{l-n+k}y^{m-k}]
\sum_{i=0}^m\binom{m}{i}x^i\sum_{j=0}^l\binom{l}{j}y^j\\
&=\sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k}
\end{align*}
In order to consider the RHS of (6), we do a little trick. We set $x=xz$ and $y=yz$ and note that
\begin{align*}
[x^0y^0]f(x,y)&=[x^0y^0z^0]f(xz,yz)\\
&=[x^0y^0z^0]\frac{(xz-yz)^n}{(xz)^l(yz)^m(1-z^2xy)^{n+1}}\\
&=[x^0y^0z^0]\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\tag{7}
\end{align*}
We observe that (7) is zero if $l+m-n$ is odd. If $l+m-n=2r$ is even, we obtain
\begin{align*}
[x^0y^0z^0]&\frac{(x-y)^n}{x^ly^mz^{l+m-n}(1-z^2xy)^{n+1}}\\
&=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{-(n+1)}{j}(-z^2xy)^j\\
&=[x^ly^mz^{2r}](x-y)^n\sum_{j\geq 0}\binom{n+j}{j}(z^2xy)^j\\
&=\binom{n+r}{r}[x^{l-r}y^{m-r}]\sum_{j=0}^nx^k(-y)^{n-k}\\
&=\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{8}
\end{align*}

With focus on (8) at which $l+m-n=2r$ is even we conclude  
\begin{align*}
\sum_{k}(-1)^k\binom{n}{k}\binom{m}{l-n+k}\binom{l}{l-m+k}
=\binom{n+r}{r}\binom{n}{l-r}(-1)^{m-r}\tag{9}
\end{align*}
Step 3: We can now put more symmetry into (9) by setting
\begin{align*}
n&=a+b\\
m&=c+a\\
l&=b+c
\end{align*}
  We obtain
  \begin{align*}
\sum_{k}(-1)^k\binom{a+b}{k}\binom{c+a}{c-a+k}\binom{b+c}{b-a+k}=(-1)^{a}\binom{a+b+c}{a+b}\binom{a+b}{a}
\end{align*}
  Finally multiplying this identity with $(-1)^a$ and shifting the index $k\rightarrow k+a$ we obtain
  \begin{align*}
\sum_{k}(-1)^k\binom{a+b}{a+k}\binom{b+c}{b+k}\binom{c+a}{c+k}&=\binom{a+b+c}{a+b}\binom{a+b}{a}\\
&=\frac{(a+b+c)!}{a!b!c!}
\end{align*}
  and the claim follows.

</s>