Question

if x+y+z =0 ,prove that x^3+y^3+z^3=3xyz

Answer 1

this is the answer hope hope It helped you

Answer 2

BECAUSE
IN IDENTITY
$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = (x + y + z)(x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - zx)$
if x+y+z=0
then
$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = (0)(x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - zx)$
then
$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = 0$
then
$x {}^{3} + y {}^{3} + z {}^{3} = 3xyz$
HENCE PROVED
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