Find $Gal(K/\mathbb{Q})$ and show that $K/\mathbb{Q}$ is normal where $K=\mathbb{Q}(a)$ 
Let $K=\mathbb{Q}(a)$ and $a$ is a root of $x^3+x^2-2x-1 \in \mathbb{Q}[x]$. Find $Gal(K/\mathbb{Q})$ and prove that $K/\mathbb{Q}$ is normal.

I just noticed that $a^2-2$ is also a root of the polynomial. But I don't know how to use that to find the Galois group. To show that $K/\mathbb{Q}$ is normal, do I need to show that every polynomial in $\mathbb{Q}[x]$ with at least one root in $K$ splits completely in $K$? If so, how do I prove it with given information?
 A: Let $f(x)=x^3+x^2-2x-1$. Since the degree of $f$ is $3$, we know that $f$ has a real root $a$. Suppose $a\in Q$. Write $a={p\over q}$ where $gcd(p,q)=1$.
We have ${p^3\over q^3}+{p^2\over q^2}-2{p\over q}-1=0$. This implies that
$p^3+qp^2-2q^2p=p(p^2+qp-2q^2)=q^3$. This implies $p$ divides $q^3$ and $p=1$ or $-1$ since $gcd(p,q)=1$. We can also write $p^3=q(q^2-p^2+2qp)$ thus $q$ divides $p^3$ and henceforth $q=1$ or $-1$. This implies that $a=1$ or $a=-1$. But obviously $-1$ and $1$ are not roots of $f$. Thus $a$ is not in $Q$.
The minimal polynomial $P$ of $a$ divides $f$. Its degree is not $1$ since $a$ is not in $Q$. Suppose that the degree of $a$ is $2$, then ${f\over P}$ is a degree $1$ polynomial which has a root in $Q$ impossible, since this root is also a root of $f$. Thus $P=f$ and $K=Q[x]/(f)$. The orbit of $a$ by $G$ the Galois group of $[Q(a):Q]$ is $1,2$ or $3$. It can't be 1 or 2 since the degree of the minimal polynomial of $a$ is $3$ so it is $3$. This implies that $f=(x-a)(x-s(a))(x-t(a)), s,t\in Gal(Q(a):Q)=Z/3$. 
A: A standard way of doing this would be: 
First check that the cubic polynomial
$$ f(x) = x^3+x^2-2x-1$$
is irreducible. If $f(x)$ weren't irreducible, since it is a cubic, there would be a rational root. However, the only possible rational roots are $1$ or $-1$, and $f(1)\not=0$ and $f(-1)\not = 0$. So $f$ is irreducible.
Therefore the Galois group is either $S_3$ or $A_3$, the only transitive subgroups of $S_3$. (A separable poly is irreducible iff the Galois group acts transitively on the roots.)
To distinguish between the two, calculate the discriminant: Wolfram Alpha (http://www.wolframalpha.com/input/?i=discriminant+(x%5E3%2Bx%5E2-2x-1+)+) tells us that it is a rational square. 
Therefore, we are done: the Galois group is the alternating group $A_3\simeq \mathbb Z/3$, and $K$ is the splitting field of $f(x)$ (and thus normal).  
