If $x + 2 y + 3 z = 0$ then $x^{3} + 8 y^{3} + 27 z^{3}$ = _____________

$18 x y z$

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$x y z$

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$3 x y z$

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$0$

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Solution

The correct option is A
$18 x y z$

Given that $x + 2 y + 3 z = 0$
Hence $x + 2 y = - 3 z$
Taking cube on both sides, we have
$(x + 2 y)^{3} = (- 3 z))^{3}$
$\Rightarrow (x)^{3} + (2 y)^{3} + (3) (x) (2 y) (x + 2 y) = (- 27 z^{3})$
$x + 2 y = - 3 z$
$\Rightarrow (x)^{3} + (2 y)^{3} + (3) (x) (2 y) (- 3 z) = (- 27 z^{3})$
$\Rightarrow (x^{3} + 8 y^{3} + (3) (x) (2 y) (- 3 z) = (- 27 z^{3})$

$\Rightarrow (x^{3} + 8 y^{3} + 27 z^{3}) = (3) (x) (2 y) (3 z)$
$= 18 x y z$

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