Question

prove that x+y+z = 0. then
x^3+y^3+z^3= 3xyz​

Answer 1

Answer:

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Answer 2

$x + y + z = 0 = > x + y = - z \\ = > (x + y)^{3} = ( - z )^{3} \\ = > x^{3} + y ^{3} + 3xy \: (x + y) = ( - z) = - {z} \\ = > x^{3} + y ^{3} + 3xy ( - z ) = - {z}^{3} \\ = > x^{3} + y ^{3} - 3xyz = 3xyz$

$Hence,(x + y + z) = 0 = > ( {x}^{3} {y}^{3} {z}^{ 3} ) = 3xyz$