When will $a \pm b\sqrt{c}$ will have a nth root of the same form? So recently I had a homework problem for my abstract algebra class which I solved where I had to prove $\sqrt[3]{8+\sqrt{325}} +\sqrt[3]{8-\sqrt{325}}=3$.  It was simple enough, all one has to do is cube it and after a bit of algebra note the cubic polynomial is relatively easy to factor.
I did not do this though, instead my approach was a bit more complicated (I missed the simple approach).  I made an educated guess that the form of the cube root of $8 \pm \sqrt{325}$ would be $m \pm n\sqrt{13}$.  This was because of at least a couple of things I had noted.  Similar already solved problems in the book had solutions essentially in this form.  Moreover, I knew I needed it to simplify to a rational number so I really wanted the squares to cancel out, and intuitively the form just made sense to me.
Yet while my intuition got me the right answer, I really feel like something deeper is lurking here.  Part of me suspects that its possible to prove with some conditions imposed on a,b,c that the nth root of $a \pm b\sqrt{c}$ must also be of that form, though I suspect some weirdness potentially (and I am unsure how to exactly state this).  Though, a large part of me is very unsure.  Why should I even suspect roots to exist in that field, $\mathbb Q[\sqrt{13}]$ is obviously not algebraically closed since -1 doesn't have a root.
So, with $a,b,c \in \mathbb Q$ when will $a \pm b\sqrt{c}$ have an nth root of the same form?
 A: There is a naive algorithm:
Wlog $c$ is a non-square integer and $a,b$ are integers.
$K=\Bbb{Q}(\sqrt{c})$ is a quadratic extension of $\Bbb{Q}$,
$O_K \subset \frac1{4c}\Bbb{Z}[\sqrt{c}]$,
If $x^n-a-b\sqrt{c}$ has a root $\mu$ in $K$ then it is in $O_K$, ie. $$\mu=\frac{r+s\sqrt{c}}{4c},\quad r,s\in \Bbb{Z}$$
The roots of $x^n-a\pm b\sqrt{c}$ have absolute value $\le M=(|a|+|b|\sqrt{|c|})^{1/n}$.
This implies that both $|\frac{r+s\sqrt{c}}{4c}|,|\frac{r-s\sqrt{c}}{4c}|\le M$.
ie. $|r|\le M 4c,|s|\le M4\sqrt{|c|}$, which means finitely elements of $K$ to check.
A: The example you were asked comes from the auxiliary quadratic associated to a cubic, so is a special case.
Take a cubic $f=(X-a)(X^2+bX+c)$. Set $d=(b^2-4c)$, the discriminant of the quadratic. Then the discriminant of $f$ is $\Delta=(a^2+ab+c)^2d$, the auxiliary quadratic has discriminant $-27\Delta$ and roots $u^3,v^3$, and writing $\omega\in\mathbb Q(\sqrt{-3})$ we have that
$$\mathbb Q(u,v,\omega)=\mathbb Q(\sqrt{-3},\sqrt d).$$
Thus $u^3,v^3\in\mathbb Q(\sqrt{-3d})$ and have cube roots in $\mathbb Q(\sqrt{-3},\sqrt d)$. The only possibility is then that $u,v\in\mathbb Q(\sqrt{-3d})$ (since a polynomial $X^3-\alpha$ is either irreducible or has a root. This is included for example in the online notes by K. Conrad on simple radical extensions).
In summary, every such situation yields elements in a quadratic extension having cube roots in that same extension.
A: Say you are looking for the $n$-th root of $a+ b\sqrt{c}$ to be a quadratic irrational. Consider the equation of $\sqrt[n]{a+b\sqrt{c}}$, which is
$$x^{2n} - 2 a x^n + (a^2 - b^2 c)$$
and search for a quadratic factor of it. Then you'll find your root.
Example. Take $\alpha = 2762 - 769 \sqrt{13}$. To find its fifth root as a quadratic irrational.
Solution:

*

*Set up the quadratic equation for $\alpha$
$$P(x) = x^2 - 2 \cdot 2762 x + (2762^2 - 769^2 \cdot 13) = x^2 - 5524 x - 59049$$


*Factor the polynomial  $P(x^5)$
$$x^{10} - 5524 x^5 - 59049= (x^2 - 4 x - 9) \times Q(x)$$


*One of the roots of $x^2 - 4 x - 9$ will be your solution.

Now we see a semblance of condition : a certain polynomial must have a quadratic factor. Is that easy to transform into another condition, I am not sure.
