Question

If x+y +z=0 then prove that x 3 +y 3 +z 3 = 3xyz

Answer 1

x+y+z=0

=x+y=-z

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

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

But x+y=-z

x^3+y^3-3xyz+z^3=0

=x^3+y^3+z^3=3xyz[Proved]

Answer 2

Heya mate, Here is ur answer

x+y+z=0

x+y=-z

cubing both the sides

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

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

but x+y=-z

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

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

x^3 +Y^3+ z^3 = 3xyz

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@Laughterqueen

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