<s>[INST] Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem 
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an useful property:

If $a+b+c=0$ then $a^3+b^3+c^3=3abc$

It can be easily proved using the following factorization
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
I tried using this one for the rationalization. I got 
$$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}*\frac{(a^2+b^2+c^2-ab-bc-ac)}{(a^2+b^2+c^2-ab-bc-ac)}=\frac{\sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}-\sqrt[3]{ab}-\sqrt[3]{bc}-\sqrt[3]{ac}}{a+b+c-3\sqrt[3]{abc}}$$
But i didn't know how to proceed. I tried looking into it with wolfram alpha and i got there is no racionalization, so i assume $a+b+c=0 $ (not $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$, because the first expression wouldn't have sense) if we want a solution (In fact, if this happens, i've already rationalized it). 
So my truly question is not the answer, but how to prove that $a+b+c$ needs to be 0 
I would appreciate any help, ideas or suggestions, thanks.
 [/INST] Maple rationalizes it as 
$$ -{\frac { \left( {x}^{2/3}+2\,\sqrt [3]{x}\sqrt [3]{y}-\sqrt [3]{z}
\sqrt [3]{x}+{y}^{2/3}-\sqrt [3]{z}\sqrt [3]{y}+{z}^{2/3} \right) 
 \left( 3\,{x}^{5/3}\sqrt [3]{y}-6\,{x}^{2/3}{y}^{4/3}+3\,{x}^{2/3}
\sqrt [3]{y}z-6\,{x}^{4/3}{y}^{2/3}+3\,\sqrt [3]{x}{y}^{5/3}+3\,\sqrt 
[3]{x}{y}^{2/3}z-{x}^{2}+7\,xy-2\,xz-{y}^{2}-2\,yz-{z}^{2} \right) }{{
x}^{3}+3\,{x}^{2}y+3\,{x}^{2}z+3\,x{y}^{2}-21\,xyz+3\,x{z}^{2}+{y}^{3}
+3\,{y}^{2}z+3\,y{z}^{2}+{z}^{3}}}
$$
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