I don't know how to solve the following:
Let $\alpha$ be a real root of $f(x)=x^4+3x-3\in Q[x]$. Is $\alpha$ a constructible number?
Any help is welcome.
I tried to get some info about the Galois group $G$ of $f$ (over $\mathbb Q$). It turns out that $f$ has a root $-1$ in $\mathbb F_5$ and $f(x)/(x+1)$ is irreducible in $\mathbb F_5[x]$. The group $G\subset S_4$ thus contains a $3$-cycle (Frobenius at $p=5$), in particular the order of $G$ is not a power of $2$, so the roots of $f$ are not constructible.
As has been proven, for a quartic's solutions to be constructible, the roots of the resolvent cubic must be rational. Given this equation, the cubic coefficient of its resolvent is $-8$ and its resolvent constant coefficient is $9$. All rational solutions must be of the form $p/q$, where $p$ and $q$ are coprime integers, $p$ is a factor of 9 and $q$ is a factor of 8 (either $p$ and/or $q$ may be negative.)
This equation's resolvent is $-8z^3-(24i\sqrt{3})z^2+48z+9=0$. We can see that there are no rational solutions, because the imaginary term in the cubic's quadratic coefficient can not vanish with any real (let alone rational) nonzero $z$.