Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ 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.

 A: 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.
A: 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.
A: 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 $(5)$.
