As has been mentioned this is simply solved since it denests already in the field generated by the radicand. Generally this is not true, but there is a general denesting structure theorem that applies. Here's an extract from my Sep 15 post on denesting radicals.
DENESTING STRUCTURE THEOREM$\;\; \;$ Let $\rm\; F \;$ be a real field and
$\rm\; F' = F(q_1^{1/d1},\ldots,q_k^{1/dk}) \;$ be a real radical extension of $\rm\; F \;$
of degree $\rm\; n \;$. By $\rm\; B = \{b_0,\ldots, b_{n-1}\}$ denote the standard
basis of $\rm\; F' \;$ over $\rm\; F \;$. If $\rm\; r \;$ is in $\rm\; F' \;$ and $\rm\; d \;$ is a positive integer such
that $\rm\; r^{1/d} \;$ denests over $\rm\; F \;$ using only real radicals, that is,
$\rm\; r^{1/d} \;$ is in $\rm\; F(a_1^{1/t_1},\ldots,a_m^{1/t_m}) \;$ for some positive integers
$\rm\; t_i \;$ and positive $\rm\; a_i \in F \;$, then there exists a nonzero $\rm\; q \in F \;$ and a
$\rm\; b \in B \;$ such that $\rm\; (q b r)^{1/d} \in F' \;$.
I.e. multiplying the radicand by a $\rm\; q \;$ in the base field $\rm\; F \;$
and a power product $\rm\; b \;=\; q_1^{e_1/d_1}\cdots q_k^{e_k/d_k} \;$ we can
normalize any denesting so that it denests in the field defined
by the radicand. E.g.
$$ \sqrt{\sqrt[3]5 - \sqrt[3]4} \;\;=\; \frac{1}3 (\sqrt[3]2 + \sqrt[3]{20} - \sqrt[3]{25})$$
normalises to $$ \sqrt{18\ (\sqrt[3]10 - 2)} \;\;=\; 2 + 2\ \sqrt[3]{10} - \sqrt[3]{10}^2 $$
An example with nontrivial $\rm\:b$
$$ \sqrt{12 + 5\ \sqrt 6} \;\;=\; (\sqrt 2 + \sqrt 3)\ 6^{1/4} $$
normalises to
$$ \sqrt{\frac{1}3 \sqrt{6}\: (12 + 5\ \sqrt 6)} \;\;=\; 2 + \sqrt{6} $$
See said post for further details and references.
There do exist general denesting algorithms employing Galois theory, but for the simple case of quadratic algebraic numbers we can employ a simple rule that I discovered as a teenager.
Simple Denesting Rule $\rm \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\, b^2 $
and, $ $ furthermore, $\rm\ w\:$ has $ $ trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2a$
Here $\:61-24\sqrt{5}\:$ has norm $= 29^2.\:$ $\rm\, \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 29\ $ yields $\ 32-24\sqrt{5}\:$
and this has $\rm\ \sqrt{trace}\: =\: 8,\ \ thus,\ \ \ \color{brown}{dividing \ it \ out}\, $ of this yields the sqrt: $\,\pm( 4\,-\,3\sqrt{5}).$
See here for a simple proof of the rule, and see here for many examples of its use.
Best Answer
For nested cubic-roots of the form $\sqrt[3]{p \sqrt[3]{q}+r}$, there is a systematic denesting method due to Ramanujan.
If the cubic polynomial $x^3 +a x^2 +bx +c=(x-x_1)(x-x_2)(x-x_3)$ satisfies
$$b +a c^{1/3} +3c^{2/3}=0\tag1 $$
per Ramanujan $$\sqrt[3]{x_1}+ \sqrt[3]{x_2 }+ \sqrt[3]{x_3 }=\sqrt[3]{ 3\sqrt[3]{9c-ab }-a-6\sqrt[3]c} \tag2$$
To denest $\sqrt[3]{\sqrt[3]{2}-1}$, match it to the right-hand-side of (2)
$$a+6c^{1/3}=1,\>\>\>\>\>27(9c-ab)=2$$
which, together with (1), determines $a=-\frac13$, $b=-\frac2{27}$ and $c= \frac8{729}$, or
$$x^3 -\frac13 x^2-\frac2{27}x + \frac8{729}= (x-\frac19)(x+\frac29)(x-\frac49)$$
Then, the Ramanujan’s formula (2) yields
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$