Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$

Question

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.

Answer 1

Note that $$(x+y+z)^3xyz-(xz+xy+yz)^3=(x^2-yz)(y^2-xz)(z^2-xy)$$ and

$$(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3=3 (b-a) (c-a) (a-x) (b-c) (b-x) (c-x)$$

Answer 2

And

$$ (x - a)^3 (b - c)^3 + (x - b)^3 (c - a)^3 + (x - c)^3 (a - b)^3 = 3 (a - b) (a - c) (b - c) (a - x) (b - x) (c - x) $$

This is easy to conclude because $a,b,c$ are roots

Answer 3

Let $$\, A(x,y,z) := (x+y+z)^3xyz-(xz+xy+yz)^3 .\,$$ Then $$\, A(x,xt,x/u) = x^6/u^4 ( t(1+u+tu)^3 - u(1+t+tu)^3 ). \,$$ That quantity is clearly $\,0\,$ if $\, t=u \,$ or $\, x^2 = xt\, x/u = yz. \,$ or $\, x^2 - yz = 0.\,$ Do the same thing with $\, y \,$ and $\, z \,$ to get the three factors $\, A(x,y,z) = c(x^2-yz)(y^2-xz)(z^2-xy). \,$ The constant $\, c=1.$

Answer 4

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 $\,z^2-xy\,$.