Factorize $(a+b+c)^5-(a+b-c)^5-(a+c-b)^5-(b+c-a)^5$ We can see that $a=0$ makes expression $0$, similarly others make expression $0$.This implies that its factorization has $abc$ in it.I substituted $a+b+c=s$ tried to find remaining factors.From here I did not progress any further
Later I went on finding this on wolframalpha and it's factorization is $80abc(a^2+b^2+c^2)$. How can we get its factorization by hand without actually expanding whole thing?Is there any slick way?

(Along Angina Seng's answer)
The given polynomial $p$ is homogeneous of degree $5$ and symmetric in $a$, $b$, $c$. Furthermore
$$p=\bigl((a+b)+c)\bigr)^5-\bigl((a+b)-c\bigr)^5-\bigl((a-b)+c\bigr)^5+\bigl((a-b)-c\bigr)^5\ ,$$
hence $p$ is odd in each variable. It follows that $p=abc\ Q(a,b,c)$, where $Q$ is a symmetric homogeneous polynomial of degree $2$ which is even in each variable. It follows that $Q(a,b,c)=N(a^2+b^2+c^2)$ for a certain constant $N$. Letting $a=b=c=1$ gives $N=80$.
