Exercise NO. 1 : Algebra, x+y+z=0


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


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/


2 Responses to “Exercise NO. 1 : Algebra, x+y+z=0”

  1. 1 Ofir

    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)

  2. 2 masteranza

    Very cool. Thank you Ofir

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