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The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.

Ramanujan established many such denesting of radicals such as $$\sqrt{\sqrt[5]{\frac{1}{5}} + \sqrt[5]{\frac{4}{5}}} = \sqrt[5]{1 + \sqrt[5]{2} + \sqrt[5]{8}} = \sqrt[5]{\frac{16}{125}} + \sqrt[5]{\frac{8}{125}} + \sqrt[5]{\frac{2}{125}} - \sqrt[5]{\frac{1}{125}}\tag {2}$$$$\sqrt[3]{\sqrt[5]{\frac{32}{5}} - \sqrt[5]{\frac{27}{5}}} = \sqrt[5]{\frac{1}{25}} + \sqrt[5]{\frac{3}{25}} - \sqrt[5]{\frac{9}{25}}\tag {3}$$$$\sqrt[4]{\frac{3 + 2\sqrt[4]{5}}{3 - 2\sqrt[4]{5}}} = \frac{\sqrt[4]{5} + 1}{\sqrt[4]{5} - 1}\tag{4}$$$$\sqrt[\color{red}6]{7\sqrt[3]{20} - 19} = \sqrt[3]{\frac{5}{3}} - \sqrt[3]{\frac{2}{3}}\tag{5}$$$$\sqrt[6]{4\sqrt[3]{\frac{2}{3}} - 5\sqrt[3]{\frac{1}{3}}} = \sqrt[3]{\frac{4}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{1}{9}}\tag{6}$$

$$\sqrt[8]{1\pm\sqrt{1-\left(\frac{-1+\sqrt{5}}{2}\right)^{24}}} = \frac{-1+\sqrt{5}}{2}\,\frac{\sqrt[4]{5}\pm 1}{\sqrt{2}}\tag{7}$$

with the last one found in Ramanujan's Notebooks, Vol 5, p. 300. Most of these radical expressions are units (a unit is an algebraic integer $\alpha$ such that $\alpha\beta = 1$ where $\beta$ is another algebraic integer).

For me the only way to establish these identities is to raise each side of the equation to an appropriate power using brute force algebra and then check the equality. However for higher powers (for example equation $(2)$ above) this seems very difficult.

Is there any underlying structure in these powers of units which gives rise to such identities or these are mere strange cases which were noticed by Ramanujan who used to play with all sorts of numbers as a sort of hobby? I believe (though not certain) that perhaps Ramanujan did have some idea of such structure which leads to some really nice relationships between units and their powers. I wonder if there is any sound theory of such relationships which can be exploited to give many such identities between nested and denested radicals.

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    $\begingroup$ Yes, there is an algorithm which essentially comes from Kummer theory. See this. I think I am going to bet that Ramanujan didn't know about Galois theory. If he did, he might have just discovered algebraic number theory from it's roots (ps : Ramanujan was indeed the founder of algebraic number theory, but in a different sense. He actually discovered the notion of modular forms, which were formalized decades later and used frequently in algebraic number theory) $\endgroup$ – Balarka Sen Jul 19 '14 at 12:02
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  • $\begingroup$ @BalarkaSen: the structure theorem mentioned in your link seems to be what i wanted. Need to study in more detail on this theorem. thanks $\endgroup$ – Paramanand Singh Jul 19 '14 at 12:42
  • $\begingroup$ @GerryMyerson: thanks for those links to other questions. one of them contains the kummer's theory mentioned in Balarka Sen's comments. I should have put more effort in searching on MSE first before my post. $\endgroup$ – Paramanand Singh Jul 19 '14 at 12:44
  • $\begingroup$ @ParamanandSingh: I corrected a small typo for eq. 5 (in red), to be consistent with this post $\endgroup$ – Tito Piezas III Apr 14 '15 at 3:04
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We also have the following identity,

$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$

For $m=n=1$ we get $(1)$.

For $m=4$ and $n=1$ we get $(5)$.

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    $\begingroup$ +1 Very nice. Note that the LHS can be simplified as $$\sqrt[3]{(m-n)^3+9m^2n-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}$$ $\endgroup$ – Tito Piezas III Mar 26 '16 at 4:57
  • $\begingroup$ @Frank your calculation. $\endgroup$ – Michael Rozenberg May 9 '16 at 3:19
  • $\begingroup$ Can I have a proof for this identity? A possible link or a book? $\endgroup$ – Pragyaditya Das Apr 19 '17 at 8:24
  • $\begingroup$ @MichaelRozenberg Is there any algebraic proof for this? $\endgroup$ – Pragyaditya Das Apr 19 '17 at 8:25
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    $\begingroup$ @Pragyaditya Das Yes, of course! Let $a$, $b$ and $c$ be real roots of the equation $x^3+px^2+qx+r=0$. Also Let $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c=A$ and $\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ca}=B$. Hence, try to find $p$, $q$ and $r$ such that $A^3=kAB+n$ and $B^3=mAB+l$ solve the equation $x^3=(kx+n)(mx+l)$. $\endgroup$ – Michael Rozenberg Apr 19 '17 at 10:17
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I found a PDF, where the authors have working algorithms to denest nested radicals like that of Ramanujan's, without the use of Galois theory.

https://www-old.cs.uni-paderborn.de/uploads/tx_sibibtex/DenestRamanujansNestedRadicals.pdf

I think this might help in offering an alternative way, regarding relations in the form of algorithms, to simplify and denest nested radicals as opposed to the helpful links given in the comments to your question. It is certainly detailed and interesting, so I hope it's of some use to you.

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A general identity: $$\sum_{k=0}^{3n-1}\frac{(-2)^{k/3}}{9^{1/3}} = \frac{1-(-2)^n}{3}\sqrt[3]{\sqrt[3]2-1}$$ so Ramanujan's case was $n=1$, yielding the famous $$\sqrt[3]{\sqrt[3]2-1}=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}$$ and possibly sheds light on how he found other like $(2)$, $(3)$, $(5)$ and $(6)$. There are other denestations like $$(1)=\sqrt[8]{4\sqrt[3]{\frac 23}-5\sqrt[3]{\frac 13}}=\sqrt[15]{19-5\sqrt[3]2-8\sqrt[3]4}$$ and very similarly, $$\sqrt[3]{\frac 19}+2\sqrt[3]{\frac 29}-2\sqrt[3]{\frac 49}=\sqrt{4\sqrt[3]{\frac 23}-5\sqrt[3]{\frac 13}}\tag{$\star$}$$ and that $$\sqrt[4]{(\star)}=\sqrt[5]{\sqrt[3]9-\sqrt[3]{\frac 23}-\sqrt[3]{\frac 43}}=\sqrt[6]{1-2\sqrt[3]2+\sqrt[3]4}$$

The fact that we predominantly have the numerators in geometric progression, for some reason (throughout my experimentation), seems to make radicals combine and cancel out very elegantly for abnormally high powers.


Regarding $(2)$, by taking out the numerators, I noticed they can be factored when arranged as an alternating sum. Id est, $$\sqrt[5]1-\sqrt[5]2+\sqrt[5]8-\sqrt[5]{16}=(\sqrt[5]2+1)(\sqrt[5]2-1)(1-\sqrt[5]2+\sqrt[5]4)$$ Not sure if this adds anything, though.

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    $\begingroup$ Very nice examples of denesting radicals. +1 $\endgroup$ – Paramanand Singh May 29 '20 at 14:51
  • $\begingroup$ @ParamanandSingh thanks for the upvote! And yes, indeed there are many. I have found a few myself through generalising radical forms with two parameters and solving for those. It helped me uncover $$\cfrac{\sqrt[4]3+1}{\sqrt[4]3-1}=\sqrt[3]{\cfrac{\sqrt[4]27+4-\sqrt 3}{\sqrt[4]27-4+\sqrt 3}}$$ but thus far, I only get lucky to find even one or perhaps two good denestations of my own (ignoring corollaries). I don't think Ramanujan's case was a matter of luck, and the fact he found so many - by hand - convinces me he had a systematic approach, whatever it was :P $\endgroup$ – Mr Pie May 30 '20 at 4:46
  • $\begingroup$ it is a great misfortune was everyone that Ramanujan did not have the resources to write his methods. He used the costly paper to record final results. Most of his results are based on deep and beautiful theories. $\endgroup$ – Paramanand Singh May 30 '20 at 4:57
  • $\begingroup$ @ParamanandSingh it is also a misfortune that a lot of his most powerful theories on integration, combinatorial analysis, and infinite series and inversions. But on his theories about mock-theta functions, modular forms, partitions, continued fractions, radical denestations, incredible stuff. Even his proof of Bertrand's postulate is impeccable, often called a calculus-free "bare hands" derivation, in which amidst such brevity he even deduces Chebyshev-type upper and lower bounds for the prime counting function. Just faultless! $\endgroup$ – Mr Pie May 30 '20 at 6:09

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