Factor Equation

Question

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.

Answer 1

$x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$

$=x^3(y-z)+yz(y^2-z^2)-x(y^3-z^3)$

$=x^3(y-z)+yz(y+z)(y-z)-x(y-z)(y^2+yz+z^2)$

$=(y-z)\{x^3+yz(y+z)-x(y^2+yz+z^2)\}$

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, $x(x+y)-yz-z^2$

$=x^2+xy-yz-z^2=(x+z)(x-z)+y(x-z)=-(z-x)(x+y+z)$

Answer 2

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 $x^3y, 1=-k\implies k=-1$

Answer 3

You were on the right track!

$x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$

$=x\left(z^3-x^3 + x^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$

$=\left(z^3-x^3\right)(x-y) + \left(x^3-y^3\right) (x - z)$

$=(z-x)(x-y)[z^2 + x^2 + zx -x^2 -y^2 -xy]$

$=(z-x)(x-y)[z^2 -y^2 + zx -xy]$

$=(z-x)(x-y)(z-y)[z + y + x]$