Hint: Why must $\text{Gal}(E/F)$ act transitively on the roots of $f$? What are the transitive subgroups of $S_3$?
Please let me know if you would like further elaboration. Hope this helps!
I seem to be giving lots of answers that depend on very special properties of the supplied example. Here’s an argument, tailored to your polynomial $f(x)=x^3-x-1$. Not the general method you were hoping for at all.
First set $\alpha$ to ba a root of your polynomial, which we all know is irreducible over $\Bbb Q$. It’s not too hard to calculate the discriminant of the ring $\Bbb Z[\alpha]$ as the norm down to $\Bbb Q$ of $f'(\alpha)=3\alpha^2-1$; this number is $23$, surprisingly small for a cubic extension. The fact that it’s square-free implies that $\Bbb Z[\alpha]$ is the ring of integers in the field $k=\Bbb Q(\alpha)$.
Our field $k$ clearly is not totally real, since $f$ has only one real root. So in the jargon of algebraic number theory, $r_1=r_2=1$, one real and one (pair of) complex embedding(s). We can apply the Minkowski Bound
$$
M_k=\sqrt{|\Delta_k|}\left(\frac4\pi\right)^{r_2}\frac{n!}{n^n}\,,
$$
which for $n=3$ gives a bound less than $2$, so that $\Bbb Z[\alpha]$ is automatically a principal ideal domain.
Let’s factor the number $23$ there: we certainly know that it’s not prime, since $23$ is ramified.
Now, we already know a number of norm $23$, necessarily a prime divisor of the integer $23$, it’s $3\alpha^2-1$, and indeed $23/(3\alpha^2-1)=4 + 9\alpha - 6\alpha^2$. But better than that, $23/(3\alpha^2-1)^2=3\alpha^2-4$. This number has norm $23$ (because the norm of $23$ itself is $23^3$). So we’ve found the complete factorization of $23$.
Now let’s look more closely at $f(x)=(x-\alpha)g(x)$, for a polynomial $g$ that we can discover by Euclidean Division to be $g(x)=x^2+\alpha x+\alpha^2-1$. And the roots of $g$ are the other roots of $f$; the Quadratic Formula tells you what they are, and the discriminant of $g$ is $\alpha^2-4(\alpha^2-1)=4-3\alpha^2$. which we already know as $-23/(3\alpha^2-1)^2$. Going back to the Quadratic Formula, our other roots are
$$
\rho,\rho'=\frac{\alpha\pm\sqrt\delta}2\>,\>\delta=\frac{-23}{(3\alpha^2-1)^2}\>,\>\sqrt\delta=\frac{\sqrt{-23}}{3\alpha^2-1}\>.
$$
And that gives you your roots of this one very special cubic polynomial in terms of one root $\alpha$ and $\sqrt{-23}$.
Best Answer
Hint
What kind of permutations do (1), (2) (separately) imply occur in $\operatorname{Gal}(f)$?