<s>[INST] 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.
 [/INST] 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.
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