Rationalize a fraction. Rationalize the denominator $\frac{1}{2^{\frac{1}{3}} + 3^{\frac{1}{3}} + 4^{\frac{1}{3}}}$. Is there a short solution for this task ? Thanks in advance.
 A: First, use the identity
$$(a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \;\; = \;\; a^3 + b^3 + c^3 - 3abc$$
with $a = \sqrt[3]{2}$ and $b = \sqrt[3]{3}$ and $c = \sqrt[3]{4}.$
Multiplying both the numerator and denominator of your fraction by the numerical form of $a^2 + b^2 + c^2 - ab - ac - bc$ will give you
$$ \frac{\text{stuff}}{2 + 3 + 4 - 3\sqrt[3]{24}} \;\; = \;\; \frac{\text{stuff}}{9 - 6\sqrt[3]{3}} $$
Now use the identity
$$(a-b)(a^2 + ab + b^2) = a^3 \, – \, b^3$$
with $a = 9$ and $b = 6\sqrt[3]{3}.$
Multiplying both the numerator and denominator of the displayed fraction above by the numerical form of $a^2 + ab + b^2$ will give you
$$ \frac{(\text{stuff})(\text{other stuff})}{9^3 - (6\sqrt[3]{3})^3} \;\; = \;\; \frac{(\text{stuff})(\text{other stuff})}{81}$$
At this point you simply need to multiply out (stuff)(other stuff). This is a bit tedious, but not really excessively so --- it'll be polynomial with $6$ terms multiplied by a polynomial with $3$ terms, for a total of $18$ "FOIL multiplications" that need to be carried out.
A: I found the minimal polynomial for $x=\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}$
$x^9-18 x^7-27 x^6+108 x^5+162 x^4-459 x^3-972 x^2-81=0$
The I divided all terms by $81x$ and I got 
$$\frac{1}{x}=\frac{x^8}{81}-\frac{2 x^6}{9}-\frac{x^5}{3}+\frac{4 x^4}{3}+2 x^3-\frac{17 x^2}{3}-12 x$$
Then I substituted $x$ with $\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}$ in the RHS getting after some trivial simplification
$$\frac{1}{\sqrt[3]{2}+\sqrt[3]{3}+\sqrt[3]{4}}=\frac{1}{9} \left(6 \sqrt[3]{2}-3\sqrt[3]{4}-3\sqrt[3]{12}+\sqrt[3]{9}+2 \sqrt[3]{18} +3 \sqrt[3]{6}-2\sqrt[3]{36}\right)$$
Hope this helps
PS
Without a CAS it is unthinkable to simplify the fraction as I did
