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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.

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  • $\begingroup$ Please use MathJax and line breaks to format your posts. $\endgroup$ – jgon Mar 25 at 22:21
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    $\begingroup$ I made the suggestion because you will find more people willing to help you with your problem the easier you make it for them to understand your problem. $\endgroup$ – jgon Mar 25 at 22:47
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If $f$ doesn't have a rational root, then $f$ is an irreducible cubic in $\Bbb{Q}[x]$. As a result, its roots all have degree $3$ over $\Bbb{Q}$, so they aren't constructible.

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  • $\begingroup$ Why does it follow that if f doesn't have a rational root, then f (cubic) is irreducible over the rationals? Is this from a particular theorem? $\endgroup$ – Sanjoy Kundu Mar 27 at 1:45
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    $\begingroup$ @SanjoyTheManjoy The reason is that if $f=gh$, with $g$ and $h$ nonconstant and $f$ of degree 3, then one of $g$ or $h$ has degree one, i.e., one of the factors of $f$ is linear when it isn't irreducible. Thus when $f$ is not irreducible, it has a root in the base field. $\endgroup$ – jgon Mar 27 at 2:31

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