Is there a cubic $\in \mathbb Q[x]$ whose Galois group over $\mathbb Q$ is cyclic of order $3$? 
Does there exist $f(x) \in \mathbb Q[x]$, $\deg f(x) = 3$, such that for $E$ being a splitting field of $f(x)$ over $\mathbb Q$, we have $\text{Gal}(E/\mathbb Q) = \mathbb Z_3$?

Motivation for the question: I am asked to prove that if the Galois group of a splitting field of a cubic over $\mathbb Q$ is cyclic of order $3$ then all the roots of this cubic are real. So I'm wondering what is an example of such a cubic to begin with. Thanks in advance.
 A: Let $p\equiv1\pmod3$ be a prime. Then $\Bbb Q(\zeta_p)/\Bbb Q$
is a Galois extension, cyclic of degree $p-1$ (where $\zeta_p=
\exp(2\pi i/p)$). This group has a quotient $C_3$, so has a cyclic
cubic subextension.
Simple example: $p=7$. Let $\alpha=\zeta+\zeta^{-1}$. As
$$\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$$
then
$$\zeta^3+\zeta^2+\zeta+1+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0.$$
This gives
$$\alpha^3+\alpha^2-2\alpha-1=0$$
so take $f(x)=x^3+x^2-2x-1$.
A: Underneath "Claim" in my post here, I show that the Galois group of any cubic polynomial in $\mathbb{Q}[x]$ is determined by its discriminant.  However, we can attack this with a more naive approach:
First, we know that the Galois group of a polynomial $f \in \mathbb{Q}[x]$ is a subgroup of the symmetric group $S_{\deg(f)}$, and when $f$ is irreducible, this subgroup is transitive$^\dagger$.  Therefore, the Galois group of any irreducible cubic polynomial must be either $S_3$ or $A_3 \cong \mathbb{Z}_3$.  If the cubic polynomial in question has complex roots, complex conjugation will be a nontrivial automorphism of the splitting field.  Since complex conjugation is an element of the Galois group of order $2$, we'll have $\text{Gal}(f) \cong S_3$ per Lagrange's theorem. 
So for irreducible cubic polynomials, we know $\text{Complex roots} \implies S_3$ Galois group.  But what of cubics with all real roots?  Will we necessarily have $\text{Gal}(f) \cong \mathbb{Z}_3$?  Is $\text{Gal}(f) \cong S_3$ still possible?  At this point, I think the discriminant argument is the easiest way to proceed.  Some hints are below:

 It is always the case that $D \in \mathbb{Q}$, and anything in $\mathbb{Q}$ is fixed by elements of $\text{Gal}(f)$.  Next, consider the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\sqrt{D}) \subset \mathbb{Q}(\sqrt{D}, \alpha)$, where $\alpha$ is one of the roots of $f$.  Note that the latter field is the splitting field of $f$ (why?).  


$\dagger$ For a proof of this fact, see Theorem 2.9(b) here.
$\ddagger$ Alternatively, note that complex conjugation, in general, is an odd permutation and thus not an element of an alternating group.  From this, we can conclude $\text{Gal}(f) \cong S_3$.
