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.
$x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$
$=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)$
$=x^3(y-z)+yz(y+z)(y-z)-x(y-z)(y^2+yz+z^2)$
$=(y-z)\{x^3+yz(y+z)-x(y^2+yz+z^2)\}$
Now, $x^3+yz(y+z)-x(y^2+yz+z^2)$
$=x^3+y^2z+yz^2-xy^2-xyz-z^2x$
$=x(x^2-y^2)-yz(x-y)-z^2(x-y)$
$=(x-y)\{x(x+y)-yz-z^2\}$
Now, $x(x+y)-yz-z^2$
$=x^2+xy-yz-z^2=(x+z)(x-z)+y(x-z)=-(z-x)(x+y+z)$
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 $x^3y, 1=-k\implies k=-1$
You were on the right track!
$x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=x\left(z^3-x^3 + x^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=\left(z^3-x^3\right)(x-y) + \left(x^3-y^3\right) (x - z)$
$=(z-x)(x-y)[z^2 + x^2 + zx -x^2 -y^2 -xy]$
$=(z-x)(x-y)[z^2 -y^2 + zx -xy]$
$=(z-x)(x-y)(z-y)[z + y + x]$
