Im stuck on the following:
Let $f(x) =x^3-x+1$ and let $\alpha$ be a real root of $f$. Prove that $$\sqrt{3} \neq a\alpha^2 +b\alpha +c$$ for any $a$, $b$, $c\in \mathbf{Q}.$
I am preparing for a prelim and this is in the ring/field theory section. Yet it looks mostly like a number theory question. I do not even know how to begin. All I can figure out is that $\alpha^3=\alpha-1.$ I also tried squaring both sides of $\sqrt{3}=a\alpha^2+b\alpha+c$ and got to $$3=a^2\alpha^4+2ab\alpha^3+(b^2+2ac)\alpha^2+2bc\alpha+c^2 =a^2\alpha(\alpha-1)+2ab(\alpha-1)+(b^2+2ac)\alpha^2+2bc\alpha+c^2$$ At which point I stopped because I didnt think a prelim would have us deal with this sort of mess.
Any hints or guidance is appreciated.
Just to finalise what's been commented. Checking irreducibility of $f$ is maybe easier than you think!
$f$ is cubic, checking irreducibility amounts to checking no rational roots exist. Were $f$ reducible over $\Bbb Q$, then by a well-known principle - $f$ is a monic integer polynomial - it would reduce into monic integer factors. In particular, the mysterious root $n$ would have to be an integer and $1=nk$ where $k$ is also integer. Of course $n=\pm1$ is forced, but you can check these aren't roots. It follows that $f$ is irreducible over $\Bbb Q$, and from there that $[\Bbb Q(\alpha):\Bbb Q]=\mathrm{deg}\,f=3$.
As $2\nmid3$ no intermediate quadratic extension can exist; $\sqrt{m}\in\Bbb Q(\alpha)\iff\sqrt{m}\in\Bbb Q$.