Question

x+y+z=0 show that x^3 + y^3+z^3=3xyz

Answer 1

we know that
x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx). = (x + y + z) (x^2 + y^2 + z^2 – xy – yz – zx).

When x+y+z=0
x^3 + y^3 + z^3 – 3xyz = (0) (x^2 + y^2 + z^2 – xy – yz – zx)
x^3 + y^3 + z^3 -3xyz = 0
x^3 + y^3 + z^3 = 3xyz(hence proved)

Answer 2

as the identity says that it is  x^3+y^3+z^3-3abc=(x+y+z)(x^2+y^2+z^2-ab-bc-ca)
therfore if x+y+z is 0 then it will be 3abc