How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$ While trying to answer this question, I got stuck showing that
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$
The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proceed.
That got me thinking ...
If you have some quadratic surd $a+b\sqrt{c}$, where $a$, $b$, and $c$ are integers, and $c$ is not a perfect square, how do you find out if that surd is the cube of some other surd, i.e. how to simplify nested cubic radicals of the form
$$\sqrt[3]{a+b\sqrt c}$$
 A: 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.
A: Here is a method outlined on page 52 under Algebra: Surds in Carr's Synopsis, the book from which Ramanujan taught himself most of his mathematical skills.

On page 73 in Miscellaneous Equations and Solutions, the author gives an example of how to solve a cubic equation of the form described in the above picture:

And lastly, on page 53, it is claimed that a different and arguably more generalized method can dete[c]t any $\sqrt[n]{A\pm B}$ for odd $n$ such that $A=a\sqrt x$ and $B=b\sqrt y$ for some $\{a,b,x,y\}\subset \mathbb Z^+$ with a supplied example, for which at least one of $x$ and $y$ are square-free.

A: If you write $(d+e \sqrt{f})^3=a+b \sqrt{c}$ and collect terms, you see $c=f$, then $d^3+3de^2=a, 3d^2e+e^3c=b$.  For integers, $e$ has to be a factor of $b$, $d$ has to be a factor of $a$ and you can just see if it works pretty easily.
A: Apply the known denesting formula
$$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$
where $t$ satisfies $t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0$. Take the example $\sqrt[3]{26+15\sqrt{3}}$ in question and solve
$$t^3-\frac{26}5t^2+9t-\frac{26}5=\frac15(t-2)(5t^2-16t+13)=0$$
which yields $t=2$ and the denestation
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$

Other examples are listed below, along with their resolvent equations
\begin{align}
\sqrt[3]{7+5\sqrt{2}}= 1+\sqrt{2}&\>\>\>\>\>\>\>
t^3-\frac{21}5t^2+6t-\frac{14}5 =0,\>\>\>t=1\\
\sqrt[3]{90-34\sqrt{7}}= 3-\sqrt{7}&\>\>\>\>\>\>\>
t^3+\frac{135}{17} t^2+21 t+\frac{315}{17}=0 ,\>\>\>t=3 \\
\sqrt[3]{\frac{99}2+\frac{59}2\sqrt{\frac52}}= 3+\sqrt{\frac52}&\>\>\>\>\>\>\>
t^3-\frac{297}{59} t^2+\frac{15}2 t-\frac{495}{118}=0 ,\>\>\>t=3 \\
\sqrt[3]{25+10\sqrt{5}}= \frac52+\frac12\sqrt{5}&\>\>\>\>\>\>\>
t^3-\frac{15}{2} t^2+15 t-\frac{25}{2}=0 ,\>\>\>t=5\\
\sqrt[3]{70-22\sqrt{7}}= \sqrt[3]{49}\left(1-\frac{\sqrt{7}}7\right)&\>\>\>\>\>\>\>
t^3+\frac{105}{11} t^2+21 t+\frac{245}{11}=0 ,\>\>\>t=7\\
\end{align}
