# 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.

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Have you thought about multiplying top and bottom by a²+ab+b² with a and b your cubed radicals? I am not saying it works, just a thought. – imranfat May 23 '13 at 16:55
Since $ab = 110592$, that term would just serve to increase the fraction by a huge amount and do little good. – Lee Sleek May 23 '13 at 17:00

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 $$-\frac{48+\left(12 \sqrt{849}-108\right)^{2/3}+\left(108+12 \sqrt{849}\right)^{2/3}}{18}.$$
...Which further simplifies to $-\frac{8 + \sqrt[3]{12\sqrt{849} + 620} + \sqrt[3]{12\sqrt{849} - 620}}{3}$. – Lee Sleek May 23 '13 at 18:29
Square each of $12\sqrt{849} \pm 108$, giving you single radicals. Then you can factor out 216 from each of them, place that in front as 6, and divide the whole fraction by six. – Lee Sleek May 23 '13 at 23:19