Question

Prove that :x+y+z =0

Answer 1

tell the full question it is not the full question

Answer 2

Hey Friend!!!
This may at first seem a little overcomplicated but the ideas behind it generalize to give formulas for all x^n+y^n+z^n where x+y+z=0. This is also not at all my work: it can be found in vladimir's answer to If x+y+z=0x+y+z=0, how do you prove that ((x2+y2+z2)/2)((x5+y5+z5)/5)((x2+y2+z2)/2)((x5+y5+z5)/5) =(x7+y7+z7)/7=(x7+y7+z7)/7?

Let's consider x, y and z as roots of a cubic polynomial. In particular, we want t^3+at+b=0 when t=x, y or z. The Elementary symmetric polynomials give b=-xyz and a=xy+xz+yz.

Now plug in x, y and z. We have

x^3+ax+b=0 (1).

y^3+ay+b=0 (2)

z^3+sz+b=0 (3)

Next add (1), (2) and (3) to get

(x^3+y^3+z^3)+a(x+y+z)+3b=0

But because x+y+z=0, this gives

(x^3+y^3+z^3)+3b=0

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

where we plugged in the value for b in the last step.

Make sure you fully understand that proof. You should be asking yourself why we can actually do this. Also make sure you have a good grasp of where the xyz came from in the last step.

This also may seem way overvcomplicated. However, we can multiply t^3+at+b=0 by t^k without changing  the fact that x, y and z are solutions. This would give t^(k+3)+at^(k+1)+bt^k=0 which when you plug in x, y and z and add the three resulting equations you get a recursive equation for every x^(k+3)+y^(k+3)+z^(k+3) and the derivation takes practically no brute force.
Hope it will help you...
Thanks!!!