How to find $x + y + z$?

Question

Q. If $x^{1/3} + y^{1/3} + z^{1/3} = 0$, then

(A) $x + y + z = 3 xyz$

(B) $x + y + z = 0 $

(C) $( x + y + z)^3= 27 xyz$

(D)$ x^3 + y^3 + z^3 = 0$

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What I've done: $(x^{1/3} + y^{1/3} + z^{1/3})^3 = 0^3$

$=> (x^{1/3} + y^{1/3})^3 + z+3(x^{1/3} + y^{1/3})z(x^{1/3} + y^{1/3}+z^{1/3})=0$

$=> (x^{1/3} + y^{1/3})^3 + z=0$

Now what??

Answer 1

use this

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

so if $a+b+c=0$,then we have $$a^3+b^3+c^3=3abc$$

then if $$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$$ $$\Longrightarrow x+y+z=3(xyz)^{\frac{1}{3}}$$ so $$(x+y+z)^3=27xyz$$ \begin{align*}&a^3+b^3+c^3-3abc \\ &=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)\\ &=[(a+b)^3+c^3]-3ab(a+b+c)\\ &=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)\\ &=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)\\ &=(a+b+c)(a^2+b^2+c^2-ab-bc-ac) \end{align*}