<s>[INST] 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
$$\sqrt[3]{a+b\sqrt c}$$
 [/INST] 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}
</s>