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?
 A: It's $abcQ(a,b,c)$ where $Q(a,b,c)$ is a symmetric degree $2$ function:
$$Q(a,b,c)=r(a^2+b^2+c^2)+s(ab+bc+ca).$$
We only need to test two values to find $r$ and $s$.
$$Q(1,1,1)=\frac{3^5-1^5-1^5-1^5}1=240$$
and
$$Q(1,1,-1)=\frac{1^5-3^5-(-1)^5-(-1)^5}{-1}=240.$$
So $3r+3s=3r-s=240$, leading to $s=0$ and $r=80$.
A: Let $a+b=x,\,a-b=y$ \begin{align*}
(a + b + c)^5 - (a + b - c)^5 - (a + c - b)^5 - (b + c - a)^5=\\
(x+c)^5+(c-x)^5-(y+c)^5+(y-c)^5=\\
c^5 + 5 c^4 x + 10 c^3 x^2 + 10 c^2 x^3 + 5 c x^4 + x^5+\\
c^5 - 5 c^4 x + 10 c^3 x^2 - 10 c^2 x^3 + 5 c x^4 - x^5+\\
-c^5 - 5 c^4 y - 10 c^3 y^2 - 10 c^2 y^3 - 5 c y^4 - y^5+\\
-c^5 + 5 c^4 y - 10 c^3 y^2 + 10 c^2 y^3 - 5 c y^4 + y^5=\\
20 c^3 (x^2-y^2)+10c(x^4-y^4)=\\
20 c^3 (x-y)(x+y)+10c(x-y)(x+y)(x^2+y^2)=\\
80 c^3 b a + 80 c b a (a^2+b^2)=\\
80 abc(c^2+a^2+b^2)
\end{align*}
A: (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 $N=80$.
