<s>[INST] How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
 [/INST] Compute the minimal polynomial of $\alpha=-27-3\cdot3^{1/3}+9\cdot3^{2/3}$:
$$
\begin{align}
\alpha&=-27-3\cdot3^{1/3}+9\cdot3^{2/3}\tag1\\
\alpha^2&=567+405\cdot3^{1/3}-477\cdot3^{2/3}\tag2\\
\alpha^3&=-81-25515\cdot3^{1/3}+16767\cdot3^{2/3}\tag3
\end{align}
$$
Since $(-3,405,-25515)\times(9,-477,16767)=-2214(1,81,2430)$, we get that
$$
\alpha^3+81\alpha^2+2430\alpha+19764=0\tag4
$$
Thus, dividing $(4)$ by $19764\alpha$, we get
$$
\begin{align}
\frac1\alpha
&=-\frac{\alpha^2+81\alpha+2430}{19764}\\
&=-\frac{45+9\cdot3^{1/3}+14\cdot3^{2/3}}{1098}\tag5
\end{align}
$$
</s>