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A possible way to denest $(2^{1/3}-1)^{1/3}$ is by first setting $x=\sqrt[3]{2}$ so$$x^3-1=1\implies x-1=\frac 1{1+x+x^2}=\frac 3{1+3x+3x^2+x^3}$$Multiply both sides by $9$ so$$9(x-1)=\left(\frac 3{1+x}\right)^3\implies\sqrt[3]{9(\sqrt[3]2-1)}=1-\sqrt[3]2+\sqrt[3]4$$However, I'm having trouble adapting this method to denesting both$$\begin{align*}\sqrt[3]{7\sqrt[3]{20}-1} & =\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}\tag1\\\sqrt[3]{7\sqrt[3]{20}-19} & =\sqrt[3]{\frac 49}-\sqrt[3]{\frac {80}9}+\sqrt[3]{\frac {25}9}\tag2\end{align*}$$I've only started on $(1)$ so far. Here's my work.

My work: Let $x^3=6860$ so $x^3-1=19^3$. Hence$$x-1=\frac {19^3}{1+x+x^2}=\frac {19^3\cdot3}{3+3x+3x^2}$$However, I'm not quite sure what to do after that. I don't see an easy relationship between three and $x$.

Perhaps you guys can help?

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    $\begingroup$ I am curious to know where have you seen this method called "Ramanujan's method" ? $\endgroup$ – Jean Marie Mar 27 '18 at 20:32
  • $\begingroup$ @JeanMarie I think I remember seeing a series of similar identities in one of the Ramanujan books and then again in a MSE answer $\endgroup$ – Crescendo Mar 27 '18 at 22:27
  • $\begingroup$ Your question testimonies of a good questionning, but you have to use more high level tools. When I see the tag "elementary number theory", I think that we are in the understatement. We need rather "advanced number/algebraic theory" : we have to use field/ring theories (keywords : Galois theory, Groebner bases...). $\endgroup$ – Jean Marie Mar 28 '18 at 7:02
  • $\begingroup$ @JeanMarie I originally didnt include the elementary number theory tag. $\endgroup$ – Crescendo Mar 28 '18 at 12:52
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HINT:

A method that proves the first equality but is not able to find a denesting.

One checks that both $\sqrt[3]{7\sqrt[3]{20}-1}$ and $\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$ are roots of the polynomial $x^9 + 3 x^6 + 3 x^3 - 6859$, a polynomial with a unique real root.

ADDED: Note that the LHS is a root of the equation $(x^3+1)^3= 7^3 \cdot 20=6860$. The polynomial function $(x^3+1)^3$ is strictly increasing so the equation has a unique solution.

Say we want to right LHS as a combination of cubic roots. Such a combination in general has a minimal polynomial of degree $3\cdot 3 \cdot 3=27$, unless there is some multiplicative combination of these roots that gives a rational number. For our examples, one notices that the product of these radicals is a rational number $$\sqrt[3]{\frac {16}9}\cdot \sqrt[3]{\frac 59}\cdot \sqrt[3]{\frac {100}9}=\frac{20}{9}$$

Therefore, in general we are looking for a RHS of form $$\sqrt[3]{a}+\sqrt[3]{b}-\frac{d}{\sqrt[3]{ab}}$$ with $a$, $b$, $d$ (positive) rational. Now we want the RHS to satisfy an equation of form $(x^3+1)^3 = 6860$ ( in general, if we tried to denest an expression of form $\sqrt[3]{\sqrt[3]{\alpha}-\beta}$ the equation should be $(x^3+\beta)^3=\alpha$). Note that in fact the RHS is of the form $$\sqrt[3]{u^2}+\sqrt[3]{v^2}- p \sqrt[3]{u v}$$ with $u$,$v$, $p$ are rational(positive) (there should be a reason for that, not entirely clear at this point). I'll leave it here for now.

ADDED: We might as well find an equation of degree $9$ with root $\sqrt[3]{a}+\sqrt[3]{b}-\frac{d}{\sqrt[3]{ab}}$. It is a polynomial with rational coefficients expressed as a product $$\prod_{k,l=0,1,2}(x-(\sqrt[3]{a}\omega^k+\sqrt[3]{b} \omega^l-\frac{d}{\sqrt[3]{ab}\omega^{k+l}}))$$ where $\omega= \exp(2\pi i/3)$.

I don't have a CAS system at the moment so I leave it like this. One should place conditions on $a$, $b$, $d$ so that it is a polynomial of form $(x^3+\beta)^3-\alpha$. The problem seems doable now. It may also appear necessary that $a$,$b$ are squares of rational numbers.

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