Rationalising factor of $x^{1/3}+y^{1/3}+z^{1/3}$

Question

I am working on an exercise from a book, that asks to show that a Rationalising factor of $x^{1/3}+y^{1/3}+z^{1/3}$ is $$(x^{2/3}+y^{2/3}+z^{2/3}-y^{1/3}z^{1/3}-z^{1/3}x^{1/3}-x^{1/3}y^{1/3})\\ \times (x^{2/3}+y^{2/3}+z^{2/3}+2y^{1/3}z^{1/3}-z^{1/3}x^{1/3}-x^{1/3}y^{1/3}) \\ \times(x^{2/3}+y^{2/3}+z^{2/3}-y^{1/3}z^{1/3}+2z^{1/3}x^{1/3}-x^{1/3}y^{1/3}) \\ \times (x^{2/3}+y^{2/3}+z^{2/3}-y^{1/3}z^{1/3}-z^{1/3}x^{1/3}+2x^{1/3}y^{1/3})$$ I have tried conjugate multiplication such as $(x^{1/3}+y^{1/3}+z^{1/3}) \times (x^{1/3}+y^{1/3}-z^{1/3})$, however, this produces a negative $z^{2/3}$ term which is not like the above. Any idea by which terms they multiplied to get the above multiplication?

Answer 1

HINT.- Since $x\to x^3$ is a bijection of $\mathbb R$ you can do the change of variables $(x,y,z)=(X^3,Y^3,Z^3)$ and try easier with the expression $$(X^2+Y^2+Z^2-ZY-ZX -XY)(X^2+Y^2+Z^2+2ZY-ZX -XY)(X^2+Y^2+Z^2-ZY+2ZX -XY)(X^2+Y^2+Z^2-ZY-ZX +2XY)$$