If x + y + z = 0, show that $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz.

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Solution

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

Also it is given x + y + z = 0

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

⇒ $x^{3}$ + $y^{3}$ + $z^{3}$ - 3xyz = 0

⇒ $x^{3}$ + $y^{3}$ + $z^{3}$ = 3xyz

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x^{3} + y^{3} + z^{3} = 3 x y z .

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x + y + z = 0

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

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