Find a normal extension over $\mathbb{Q}$ of degree 3 I've had trouble coming up with one.
I know that if I could find 
an irreducible poly $p(x)$ over $\mathbb{Q}$
which has roots $\alpha, \beta, \gamma\in Q(\alpha)$,
then $|\mathbb{Q}(\alpha) : \mathbb{Q}| $ = 3 and would be a normal extension,
as $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\beta,\gamma)$ would be a splitting field of $f$ over $\mathbb{Q}$.
However, this is a lot of conditions to find by luck...
Any help appreciated!
 A: Try to find a polynomial with discriminant $D$ that satisfies $\sqrt{D}\in\mathbb{Q}$.
Why does this help?
First, the only possibilities for the Galois group $G$ are $S_3$ and $A_3$, as Ben Millwood remarked.
Second,  every element of $G$ must fix $\sqrt{D}\in\mathbb{Q}$. But you can check that every transposition of two roots of $f$ does not fix $\sqrt{D}$. Therefore $G$ cannot contain any transposition and must be isomorphic to $A_3$.
Spoiler:

 Use $f(x) = x^3 -3x -1$

A: You may know that the Galois group of $x^n-1$ over the rationals is cyclic of order $\phi(n)$ (that's the Euler phi-function). If $\phi(n)$ is a multiple of $3$ (and it's not hard to find such $n$), then you can find a normal extension of the rationals of degree $3$ as a subfield of the splitting field of $x^n-1$. 
This is easiest to do if you don't choose $n$ any bigger than necessary. 
A: The splitting field of $x^3-7x+7$ is of degree 3 over the rationals, since the discriminant is a  rational square and the polynomial is irreducible.
