Question

hello friends, tell me if x+y+z=0 then prove that x³ +y³+z³=3xyz

Answer 1

Hello friend here's ur answer
◆●◆x^3+y^3+z^3-3xyz= (x+y+z)* (x^2+y^2+z^2-xy-xz-yz)+ 3xyz 

because you say that● x+y+z=0 ●then 


therefore
◆●◆x^3+y^3+z^3-3xyz= (0)*(x^2+y^2+z^2-xy-xz-yz)
=0

and thus•••,x^3+y^3+z^3=3xyz••

Answer 2

hey
let prove that

x+y+z=0
then x+y = -z

cubic both side
(x+y)^3 = -z^3
x^3 + y^3 + 3xy(x+y) =-z^3

put the value

x^3 + y^3 + 3xy (-z) = -z^3

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

transporting sides

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

///// proved//////