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If $x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$ then what is $(x+y+z)^{3}$ equal to?
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27
$x^{1 / 3}+y^{1 / 3}+z^{1 / 3}=0$
Given formula, $(a+b)^{3}=a^{3}+b^{3}+3 a b(A+b)$
so, $x^{1 / 3}=-\left(y^{1 / 3}+z^{1 / 3}\right)$
Raising both the to the of cube $x=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(y^{1 / 3}+z^{1 / 3}\right)\right\}$
$=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(-x^{1 / 3}\right)\right\}=-\left\{y+z-3 x^{1 / 3} y^{1 / 3} z^{1 / 3}\right\}$
$=-y-z+3 x^{1 / 3} y^{1 / 3} z^{1 / 3} \Rightarrow x+y+z=3 x^{1 / 3} y^{1 / 3} z^{1 / 3}$
or $x+y+z=3(x y z)^{1 / 3}$
$\Rightarrow(x+y+z)^{3}=27 x y z$
Given formula, $(a+b)^{3}=a^{3}+b^{3}+3 a b(A+b)$
so, $x^{1 / 3}=-\left(y^{1 / 3}+z^{1 / 3}\right)$
Raising both the to the of cube $x=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(y^{1 / 3}+z^{1 / 3}\right)\right\}$
$=-\left\{y+z+3 y^{1 / 3} z^{1 / 3}\left(-x^{1 / 3}\right)\right\}=-\left\{y+z-3 x^{1 / 3} y^{1 / 3} z^{1 / 3}\right\}$
$=-y-z+3 x^{1 / 3} y^{1 / 3} z^{1 / 3} \Rightarrow x+y+z=3 x^{1 / 3} y^{1 / 3} z^{1 / 3}$
or $x+y+z=3(x y z)^{1 / 3}$
$\Rightarrow(x+y+z)^{3}=27 x y z$
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