Question

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

Answer 1

➡HERE IS YOUR ANSWER⬇

We know that :

${x}^{3} + {y}^{3} + {z}^{3} - 3xyz \\ = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ \\ now \: \: whe n\: \: x + y + z = 0 \\ we \: \: get \\ \\ {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = \\ 0 \times ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) = 0 \\ \\ so \: \: {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0 \\ \\ = > \: {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$
(Proved)

⬆HOPE THIS HELPS YOU⬅