Question

If x+y+z=0
show x^3+y^3+z^3=3xyz​

Answer 1

$x + y + z = 0$

$x + y = - z$

$cubing \: on \: both \: sides$

$(x + y) { }^{3} = ( - z) {}^{3}$

$x {}^{3} + {y}^{3} + 3xy(x + y) = - z {}^{3}$

$x {}^{3} + {y}^{3} + 3xy( - z) = { - z}^{3}$

${x}^{3} + {y}^{3} - 3xyz = - z {}^{3}$

${x}^{3} + {y}^{3} + z {}^{3} = 3xyz$

hence proved.