Let $n$ be a positive integer and $S\_n$ be the symmetric group on $\{1,2,\ldots,n\}$.
For any $w\in S\_n$ and polynomial $f\in \mathbb{R}[x\_1,x\_2,\ldots,x\_n]$, denote $w(f)=f(x\_{w(1)},x\_{w(2)},\ldots,x\_{w(n)})$. Furthermore, define
$$\alpha(f)=\sum\limits\_{w\in S\_n}\epsilon(w)w(f),$$
where the sum is over all the $w\in S\_n$ and $\epsilon(w)$ denotes the sign of the permutation $w$. 


It is easy to see that for any $f\in \mathbb{R}[x\_1,x\_2,\ldots,x\_n]$, $\alpha(f)$ is divisible by $\prod\limits\_{1\leq i<j\leq n}(x\_i-x\_j).$
So when $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$ , we have $$\alpha(f)=a\_f\prod\limits\_{1\leq i<j\leq n}(x\_i-x\_j)$$
for some $a\_f\in \mathbb{R}$ .


Suppose $f$ is a homogeneous polynomial of degree $\frac{n(n-1)}{2}$, how to know whether $a\_f$ is $0$ or not when the expansion $$f(x\_1,x\_2,\ldots,x\_n)=\sum\limits\_{i\_1+i\_2+\cdots+i\_n=\frac{n(n-1)}{2}}a\_{i\_1,i\_2,\ldots,i\_n}x\_1^{i\_1}x\_2^{i\_2}\cdots x\_n^{i\_n}$$
is not easy to get?


For example, let $n=5$ and 
$$f(x\_1,x\_2,x\_3,x\_4,x\_5)=(x\_1+x\_2)^2(x\_2+x\_3)^2(x\_3+x\_4)^2(x\_4+x\_5)^2(x\_5+x\_1)^2,$$
then $f$ is homogeneous polynomial of degree $10$.


With the expansion of $f$ expanded by the computer, I can calculate that $a\_f$ is nonzero. But how to get this without the expansion of $f$?


Maybe it is hard to answer my question in general, so I hope someone can answer my question for the specific polynomial I gave above.



A question about homogenous polynomials of degree $\frac{n(n-1)}{2}$
As for your polynomial $f=\prod\_{i=1}^5 (x\_i+x\_{i+1})^2$, you may do the following trick. At first, you replace your polynomial to $g=\prod\_{i=1}^5 (x\_i+x\_{i+1}-4)(x\_i+x\_{i+1}-3)$. Antisymmetrizations of $f$ and $g$ are the same, since their difference has degree less than 10. Next, we use points 0,1,2,3,4 as $x$'s. I think, there is unique (up to dihedral group symmetry) permutation 10234 for which $g$ is non-zero. Hence all non-zero summands in the sum $\sum {\rm\,sign}\,(\pi) g(x\_{\pi\_1},x\_{\pi\_2},\dots,x\_{\pi\_5})$ are equal, so this sum is non-zero.


