# Algebraic manipulation

• September 28th 2009, 04:30 PM
I-Think
Algebraic manipulation
Fairly challenging question here. I'll still try to solve it while I get the forum's help

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

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

Thanks.
• September 28th 2009, 04:48 PM
pickslides
$
\frac{x^3+y^3+z^3}{xyz}=\frac{(a(b^3-c^3))^3+(b(c^3-a^3))^3+(c(a^3-b^3))^3}{a(b^3-c^3)\times b(c^3-a^3)\times c(a^3-b^3)
} = \dots
$

Now start expanding
• September 28th 2009, 05:49 PM
I-Think
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!
• September 28th 2009, 06:15 PM
pickslides
Maybe not the only way to solve this but seems to be the most obvious.

Congrats on the 100th post. (Clapping)