TITLE: How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$
QUESTION [9 upvotes]: 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
$$\sqrt[3]{a+b\sqrt c}$$

REPLY [3 votes]: 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.