Fairly challenging question here. I'll still try to solve it while I get the forum's help

Deduce that, if$\displaystyle x=a(b^3-c^3)$,$\displaystyle y=b(c^3-a^3)$ and $\displaystyle z=c(a^3-b^3)$, then

$\displaystyle \frac{x^3+y^3+z^3}{xyz}=\frac{a^3+b^3+c^3}{abc}$

Thanks.