Prove a cubic polynomial with rational coefficients $f$ does not have a root in $Q$ implies $f$ does not have a constructible root in the reals.
My attempt at the proof goes as follows: If we suppose that f does not have a root in $Q$, then $f$ either has complex root(s) or irrational roots. More generally, we can say $f$ has a repeated real roots, $f$ has two complex conjugate roots and one real root, of $f$ has three distinct roots. What I'm having trouble showing is how to eliminate these cases. That is, I'm unable to find a way to show that one of these cases will lead to a violation of $f$ having integer coefficients. And if that is violated, then I can follow up by stating that there does not exist a chain of field extensions which would culminate with showing that $f$ does not have a constructible roots in the reals. Any help would be appreciated.