Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$. 
I am rather stuck though. Is there a general method for going about this? I always find myself having to guess which is not so useful here. I notice in the first one that if $x=y=z$ then the polynomial is zero. I'm not sure how useful this is though in the case of three variables. 

Brute-forcing the first factorization by expanding and collecting the powers of $\,x\,$:
$$
\begin{align}
(x+y+z)^3 xyz - (xz+xy+yz)^3 &= yz \,x^4 - (y^3+z^3) \,x^3 + yz(y^3+z^3)\,x - y^3 z^3 \\
 &=yz(x^4-y^2z^2)-(y^3+z^3)x(x^2-yz) \\
 &= yz(x^2-yz)(x^2+yz)-(y^3+z^3)x(x^2-yz) \\
 &= (x^2-yz)(\ldots)
\end{align}
$$
Therefore $\,x^2-yz\,$ is a factor, and by symmetry so are $\,y^2-xz\,$ and $\,z^2-xy\,$.
