### Exercise NO. 1 : Algebra, x+y+z=0

28Jan08

A few days ago I were stuck with one exercise which I can’t pushed off :] Meanwhile when I forgotten about it, friend of mine, Ecik, did it. It’s below with his solution.

Prove the following:

$x^3 + y^3 + z^3 =3xyz$

if $x+y+z=0$

Solution:

$x^3+y^3+z^3=(x+y) \cdot (x^2-xy+y^2)+z^3= \newline =-z \cdot (x^2-xy+y^2) + z^3=xyz-z \cdot (x^2+y^2-z^2) = \newline = xyz - z \cdot (x^2+y^2-(x+y)^2) = xyz - z \cdot (x^2+y^2-x^2-2xy-y^2) \newline = xyz +2xyz=3xyz$

I’m open on any suggestions or other solutions of yours.

If you’d like to see Ecik’s blog, here it is http://braindarkside.wordpress.com/

You can also use the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$