Is it possible to rationalize a denominator containing two cube roots? The fraction in question is
$$-\frac{12}{\sqrt[3]{12\sqrt{849} + 108} - \sqrt[3]{12\sqrt{849} - 108}}$$
And was reached in calculating the solution to $x^4 - x - 1 = 0$. I've tried all the standard methods, including $(a+b)(a-b) = a^2 - b^2$, but that doesn't work for cube roots, because once you have the square of one the two middle terms will not cancel each other out.

As imranfat suggests in his comment, you should use the identity
$$
\frac1{\sqrt[3]a-\sqrt[3]b} = \frac{\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}}{a-b}
$$
which can be verified via cross-multiplication. In your case, take $a=12\sqrt{849} + 108$ and $b=12\sqrt{849} - 108$ and then work through simplifying the resulting expression. When I do so, I obtain
$$
-\frac{48+\left(12 \sqrt{849}-108\right)^{2/3}+\left(108+12 \sqrt{849}\right)^{2/3}}{18}.
$$
