Factorize $ (kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) $

Question

I tried opening the brackets but it keeps getting complex and complex. Does anyone have any easier way of doing it. Please tell how did you observe.

Answer 1

$$ f(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz) \tag 1$$ Name the six terms respectively as : $$f(x,y,z,k)=A*B*C-a*b*c$$ If $k=1$ we have $A=c$ , $B=b$ and $C=b$ thus $ABC=abc$ and $f(x,y,z,1)=0$. As a consequence $(k-1)$ is a factor of $f(x,y,z,k)$.

Do the same with $k=-1$ which leads to $f(x,y,z,-1)=0$ then a second factor $(k+1)$.

Proceeding on the same manner successively with $x=y$ , $y=z$ , and $z=x$ gives three other factors : $(x-y)$ , $(y-z)$ and $(z-x)$.

Finally $$f(x,y,z,k)=-2(k-1)(k+1)(x-y)(y-z)(z-x)\tag 2$$

The coefficient $-2$ is obtained in giving an arbitrary value to each $x,y,z,k$ , of course so that $f\neq 0$ , and comparing the values of Eq.$(1)$ and Eq.$(2)$.

Answer 2

Hint Let $$P(x,y,z,k)=(kx-y+z)(x+ky-z)(x-y-kz)-(kx+y-z)(x-ky-z)(x-y+kz).$$ Prove that $$P(x,x,z,k)=P(x,y,x,k)=P(x,y,y,k)=P(x,y,z,\pm 1)=0.$$