Question

IF x+y+z=0 then prove that x³+y³+z³=3xyz

Answer 1

given
x+y+z=0. ------(1)
then
x+y= -z

cubing on both side
(x+y)^3= -z^3

formula =(x+y)^3= x^3+3xy(x+y)+ y^3

so here also
x^3+3xy(x+y)+y^3 = -z^3

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

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


proved

Answer 2

Answer:given

x+y+z=0. ------(1)

then

x+y= -z

cubing on both side

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

formula =(x+y)^3= x^3+3xy(x+y)+ y^3

so here also

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

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

hence

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

proved