Question

if (x+y+z)=0,prove that (x³+y³+z³)=3xyz

Answer 1

X+y=-z-------------(1)
Takeqube of both sides
Xqube+yqube+3xy(x+y)=-zqube
X3+y3+z3+3xy(-z) =0 (from 1)
X3+y3+z3=3xyz it is proved

Answer 2

hay!!

Dear user -

We have :

x+y+z=0

Then, =>
$x + y = - z \\ = ({x + y})^{3} =( - {z}^{3} ) \\ = {x}^{3} + {y}^{3} + 3xy(x + y) = - {z}^{3} \\ {x}^{3} + {y}^{3} + 3xy( - z) = - {z}^{3} \\ {x}^{3} + {y}^{3} - 3xyz = - {z}^{3} \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz.$
Hence, (x+y+z)=0 => (x³+y³+z³)=3xyz

I hope it's help you