Question

prove that x+y+z=0,x³+y³+z³=3xyz

Answer 1

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Here is your answer

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$x + y + z = 0 \\ x + y = - z$

Cubing both the side
${(x + y)}^{3} = { (- z)}^{3} \\ \\ {x}^{3} + {y}^{3} + 3xy(x + y) = { (- z)}^{3} \\ \\ as \: we \: solve \: that \: \\ x + y = - z \: so \\ \\ {(x)}^{3} + {(y)}^{3} + 3xy( - z) = {( -z )}^{3} \\ \\ {x}^{3} + {y}^{3} - 3xyz \: + {z}^{3} = 0 \\ \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz \: \: proved$

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Answer 2

We know that,

$x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = (x + y + z)(x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - zx)$

Given,

$x + y + z = 0$

So,

$x {}^{3} + y {}^{ 3} + z {}^{3} - 3xyz = (x + y + z)(x {}^{2} + y {}^{2} z {}^{2} - xy - yz - zx) \\ = > x {}^{3} + y {}^{3} + z {}^{3} - 3 xyz = 0 \times (x {}^{2} + y {}^{2} + z {}^{2} - xy - yz - zx) \\ = > x {}^{3} + y {}^{3} + z {}^{3} - 3xyz = 0 \\ = > x {}^{3} + y {}^{3} + z {}^{3} = 3xyz$

Hence Proved.