Question

If x+y+z=0 then prove that x3+y3+z3=3xyz by (x=y)3

Answer 1

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$x + y + z = 0 \\ \\ x + y = - z \\ \\ {(x + y)}^{3} = {( - z)}^{3} \\ \\ {x}^{3} + {y}^{3} + 3xy(x + y) = { - z}^{3} \\ \\ {x}^{3} + {y}^{3} + 3xyz = { - z}^{3} \\ \\ {x}^{3} + {y}^{3} + {z}^{3} = - 3xyz$
hope you are satisfied with my answer