Denesting Cardano's Formula For a depressed cubic equation $x^3 + px + q =0$ having exactly one real root, Cardano's formula gives the real root as $$\sqrt[3]{-\frac{q}{2} +\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{-\frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.$$ Suppose we know that the cubic equation has exactly one real root which is an integer or rational number. For example, consider $f = x^3 +x -2$ which has exactly one real root 1. Cardano's formula gives the root as $$\sqrt[3]{1 +\sqrt{\frac{28}{27}}} + \sqrt[3]{1 -\sqrt{\frac{28}{27}}}.$$ I referred to this post and used the technique mentioned there to observe that $\sqrt[3]{1 +\sqrt{\frac{28}{27}}}$ can be denested as $\frac{1}{2}+\frac{\sqrt{21}}{6}$ and $$\sqrt[3]{1 -\sqrt{\frac{28}{27}}} = \frac{1}{2}-\frac{\sqrt{21}}{6}.$$ This recovers our integral root $1,$ as desired. I tried few more examples and I was able to denest the outer cubic radical in those cases also.
This brings us to my question. Suppose we have a cubic polynomial $f \in \mathbb{Q}[x]$ which is promised to have exactly one integral/rational root and other two to be complex conjugates. Then, can we always denest the outer cubic radical as shown in the above example? Is there any counterexample and if not, what is the proof/proof idea for denesting.
I refered to this paper in search of a solution but it is heavily loaded with Galois theory which is a bit overwhelming for me at this stage. However, I did start reading the basics of Galois theory - field extensions, Galois correspondence etc. and I hope that I would be able to understand the layman terms of any answer to this question which involves Galois theory.
 A: If $n$ is your one real integer solution, then $$x^3+px+q=(x-n)(x^2+nx+b)$$ where $-nb=q, b-n^2=p$ and $n^2<4b,$ that last because there are real roots to $x^2+nx+b$ otherwise.
Substitute $p=b-n^2, q=nb$ into your formula, then see if you can eliminate the cube roots in the general case. You should be able to.
In particular:
$$\begin{align}D=\frac{q^2}{4}+\frac{p^3}{27} 
&=\frac{n^2b^2}{4} +\frac{(b-n^2)^3}{27}\\
&=\frac{27n^2b^2+4b^3-12n^2b^2+12n^4b-4n^6}{108}\\
&=\frac{4b^3+15n^2b^2+12n^4b-4n^6}{108}\\
&=\frac{(4b-n^2)(b+2n^2)^2}{108}
\end{align}$$
So  $$-\frac q2+\sqrt{D}=\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}$$
It turns out this has a simple cube root:
$$\begin{align}\frac{bn}2+\frac{ b+2n^2}{6}\sqrt{\frac{4b-n^2}{3}}&=\left(\frac{n}{2}+\frac 12\sqrt{\frac{4b-n^2}3}\right)^3
\end{align}$$
So you can always reduce the cube root.

In your case $n=1,b=2,$ and this can be written:
$$1+\frac{2}{3}\sqrt{\frac 7 3}=\left(\frac12+\frac12\sqrt{\frac73}\right)^3$$

Nothing in this result requires $n$ to be an integer. The equalities are all true regardless of what $n$ is. It even holds when there are other real roots, or if $n$ is a complex root.

For example if $p=-7,q=6,$ then $x^3-7x+6=0$ has roots $1,2,-3.$ Cardano requires us to s find the cube root of:
$$S:=\frac72 +\frac{10}{9}\sqrt{-3}$$
When $n=1, b=-6$ and you get:
$$S=\left(\frac12+\frac{1}{2}\sqrt{\frac{-25}{3}}\right)^3$$
When $n=2, b=-3,$ and
$$S=\left(1+\frac{1}{2}\sqrt{\frac{-16}{3}}\right)^3$$
When $n=-3, b=2$ and
$$S=\left(-\frac32+\frac{1}{2}\sqrt{\frac{-1}{3}}\right)^3$$
