Is it possible to rationalize a denominator containing two cube roots?

1.5k Views Asked by At

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.

1

There are 1 best solutions below

2
On BEST ANSWER

As imranfat suggests in their 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 $$ -\frac{48+\left(12 \sqrt{849}-108\right)^{2/3}+\left(108+12 \sqrt{849}\right)^{2/3}}{18}. $$