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One out of the almost endless supply of identities discovered by Ramanujan is the following: $$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{\frac19} - \sqrt[3]{\frac29} + \sqrt[3]{\frac49}, $$ which has the following interpretation in algebraic number theory: the fundamental unit $\sqrt[3]{2}-1$ of the pure cubic number field $K = {\mathbb Q}(\sqrt[3]{2})$ becomes a cube in the extension $L = K(\sqrt[3]{3})$.

Are there more examples of this kind in Ramanujan's work?

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Dear Franz, Do you really mean the cube root of $2 - 1$ (bearing in mind that $2 - 1 = 1$)? (I guess not, but I unfortunately don't have time right now to cube the right hand side and see what is actually intended.) –  Emerton Oct 24 '10 at 16:32
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Dear Matt: $2^{1/3} - 1$ is a fundamental unit, so probably that is what Franz meant to write. –  BCnrd Oct 24 '10 at 16:53
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Emerton and BCnrd: thanks - it's fixed now. –  Franz Lemmermeyer Oct 24 '10 at 17:54
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up vote 13 down vote accepted

$$(7 \sqrt[3]{20} - 19)^{1/6} = \ \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}},$$ $$\left( \frac{3 + 2 \sqrt[4]{5}}{3 - 2 \sqrt[4]{5}} \right)^{1/4}= \ \ \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1},$$ $$\left(\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}\right)^{1/2} = \ \ (1 + \sqrt[5]{2} + \sqrt[5]{8})^{1/5} = \ \ \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}},$$ and so on. Many of these were submitted by Ramanujan as problems to the Journal of the Indian Mathematical Society. See the following link: http://www.math.uiuc.edu/~berndt/jims.ps for more precise references. Quote: "although Ramanujan never used the term unit, and probably did not formally know what a unit was, he evidently realized their fundamental properties. He then recognized that taking certain powers of units often led to elegant identities."

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I would have gone with 'enery $$ $$ en.wikipedia.org/wiki/I'm_Henery_the_Eighth,_I_Am –  Will Jagy Oct 25 '10 at 1:05
    
Not a real link, in an answer rather than a comment, even if the wikipedia link has extra punctuation, one can click on the hyperlink symbol and convince MO to make the entire character string into a genuine link. –  Will Jagy Oct 25 '10 at 1:08
    
Excellent. I had no idea the song went back to 1910. –  Will Jagy Oct 25 '10 at 1:18
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