
Algebraic manipulation
Fairly challenging question here. I'll still try to solve it while I get the forum's help
Deduce that, if$\displaystyle x=a(b^3c^3)$,$\displaystyle y=b(c^3a^3)$ and $\displaystyle z=c(a^3b^3)$, then
$\displaystyle \frac{x^3+y^3+z^3}{xyz}=\frac{a^3+b^3+c^3}{abc}$
Thanks.

$\displaystyle
\frac{x^3+y^3+z^3}{xyz}=\frac{(a(b^3c^3))^3+(b(c^3a^3))^3+(c(a^3b^3))^3}{a(b^3c^3)\times b(c^3a^3)\times c(a^3b^3)
} = \dots
$
Now start expanding

Brute force and lengthy expansions
*Shudders*
I was hoping there would be a neat solution to this, but if straight expansion's the way to go then so be it.
Btw
100th post !Yay!

Maybe not the only way to solve this but seems to be the most obvious.
Congrats on the 100th post. (Clapping)