Is it possible to rationalize a denominator containing two cube roots?The fraction in question is−1212849+1083−12849−1083And was reached in calculating the solution to x4−x−1=0. I've tried all the standard methods, including (a+b)(a−b)=a2−b2, 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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Is it possible to rationalize a denominator containing two cube roots?The fraction in question is−1212849+1083−12849−1083And was reached in calculating the solution to x4−x−1=0. I&#039;ve tried all the standard methods, including (a+b)(a−b)=a2−b2, but that doesn&#039;t work for cube roots, because once you have the square of one the two middle terms will not cancel each other out.

Pedro Taylor · Accepted Answer

As imranfat suggests in his comment, you should use the identity1a3−b3=a23+ab3+b23a−bwhich can be verified via cross-multiplication. In your case, take a=12849+108 and b=12849−108 and then work through simplifying the resulting expression. When I do so, I obtain−48+(12849−108)23+(108+12849)2318.

Is it possible to rationalize a denominator containing two cube roots? The fraction in question

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