Factor Equation Help me with this,
Question: factor $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$.
Solution:
$$\begin{eqnarray}&=&x^3y-x^3z+y^3z-xy^3+xz^3-yz^3\\
&=&x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)\\
&=&x\left[(z-y)\left(z^2+zy+y^2\right)\right]+y\left[(x-z)\left(x^2+xz+z^2\right)\right]+z\left[(y-x)\left(y^2+xy+x^2\right)\right]\end{eqnarray}$$
This expression is quite simple at first glance, but I stuck up again in that line. I appreciate any help.

If $E(x,y,z)=x^3y-x^3z+y^3z-xy^3+xz^3-yz^3=x^3(y-z)+y^3(z-x)+z^3(x-y)$
Observe that $E=0$ if $x=y\implies x-y$ divides $f$
Similarly, $y-z,z-x$ divide $f$ $\implies E=(x-y)(y-z)(z-x)F$ 
As $E$ is Alternating polynomial wrt $x,y,z$
and "the ratio of two alternating polynomials is a symmetric function" (Source : wiki,  Proof Article# $4$ of this)
$F$ must be Symmetric polynomial of degree $1$ in $x,y,z$
So, $E=(x-y)(y-z)(z-x)\cdot k(x+y+z)$ where $k$ is independent of $x,y,z$
Comparing the coefficients of $x^3y, 1=-k\implies k=-1$
