For a depressed cubic equation x3+px+q=0x3+px+q=0 having exactly one real root, Cardano's formula gives the real root as 3√−q2+√q24+p327+3√−q2−√q24+p327. Suppose we know that the cubic equation has exactly one real root which is an integer or rational number. For example, consider f=x3+x−2 which has exactly one real root 1. Cardano's formula gives the root as 3√1+√2827+3√1−√2827. I referred to this post and used the technique mentioned there to observe that 3√1+√2827 can be denested as 12+√216 and 3√1−√2827=12−√216. 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∈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.
