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I've come across a very curious Ramanujan identity $$\sqrt[6]{7 \sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}$$

You could probably prove this by taking the 6th power of both sides. But how could he have come to this clean result without extra terms left over? I'm overwhelmed at trying to imagine how this would be done.

Edit: Also, how would you generalize identities involving a 6th root as two 3rd roots?

qwr
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1 Answers1

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Hint $\ $ Multiplying by $\,\sqrt[3]{12}\,$ the RHS becomes $\, \sqrt[3]{20}- 2,\,$ and

$$(\sqrt[3]{20}- 2)^6\, =\, 144\, (7 \sqrt[3]{20} -19)$$

One can prove by Galois theory a structure theorem which essentially implies that all radical denestings can similarly be "normalised" to such a "trivial" denesting. This forms the foundation for effective algorithms to compute such denestings. See my post here and here for much more, including examples and literature references.

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Bill Dubuque
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  • Thank you for your posts, they are very informative. Is there a general identity for a 6th root as two 3rd roots? – qwr Feb 02 '14 at 02:00
  • No, there is no universal identity. Instead, see the structure theorem in the linked posts. – Bill Dubuque Feb 02 '14 at 02:03