Question

if X+y+z=0 to prove that x^3+y^3+z^3=3xyz

Answer 1

The given condition is,
x+y+z =0
So, x+y = -z
Cubing both sides,
(x+y)³=(-y)³
x³+y³+3xy(x+y) = -y³
x³+y³+z³ = -3xy(x+y)
as, (x+y) equals -z 
x³+y³+z³ = -3xy*(-z)
x³+y³+z³= 3xyz.
Hope this helps.

Answer 2

Given :

x + y + z = 0

Transpose z to the other side so that we can easily prove the given statement .

x + y = - z

Now we will cube both sides :

= > ( x + y )³ = ( - z )³

We know that the formula for expansion :

( a + b )³ = a³ + b³ + 3 ab ( a + b )

Similarly the given value will be :

x³ + y³ + 3 xy ( x + y ) = - z³

We had written earlier that x + y + z = 0

This means that x + y = - z .

So x³ + y³ + 3 xy ( - z ) = - z³

= > x³ + y³ - 3 xyz = - z³

Transposing - z³ will change its sign :

= > x³ + y³ + z³ - 3 xyz = 0

Finally transpose the - 3 xyz to the other side :

= > x³ + y³ + z³ = 3 xyz .

Hence the given problem is proved !