A basis of a Galois extension. Let $f(x) = x^3 - 3x + 1 \in \mathbb{Q}[x]$. It is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$. The discriminant of $f$ is $81 = 9^2$, so the Galois group of $f$ is isomorphic to $A_3$ and the splitting field of $f$ is $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of $f$. 
Now I want to represent the elements of $G(\mathbb{Q}(\alpha)/\mathbb{Q})$ as matrices $3 \times 3$, but to do that I need a basis of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$. And the question is:

What basis of $\mathbb{Q}(\alpha)/\mathbb{Q}$ is better to choose and how to express the roots $\alpha, \beta$ and $\gamma$ in terms of that basis?

If I take ${1, \alpha, \alpha^2}$ as a basis, what is a good way of finding the coordinates of $\beta$ and $\gamma$?

The roots of $f(x)$ will be of the form $\alpha, \beta, -\alpha - \beta$, as the coefficient of $x^{2}$ is zero. Now a basis of $\mathbb{Q}(\alpha)/\mathbb{Q}$ will be
$$\tag{basis}
1, \alpha, \beta.
$$
You know that one of the generators $g$ of the Galois group will permute the roots cyclically
$$
\alpha \mapsto \beta \mapsto - \alpha - \beta \mapsto \alpha.
$$
Then, using (basis), a matrix for this generator will be
$$
\begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & -1\\
0 & 1 & -1\\
\end{bmatrix}.
$$
