TITLE: Is it possible to rationalize a denominator containing two cube roots?
QUESTION [5 upvotes]: 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.

REPLY [4 votes]: 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
$$
-\frac{48+\left(12 \sqrt{849}-108\right)^{2/3}+\left(108+12 \sqrt{849}\right)^{2/3}}{18}.
$$