Writing $2$ roots of a cubic in terms of the third root Let $\theta$ be a root of $x^3-3x+1$. Since the discriminant is a square, the splitting field of this polynomial is just $\mathbb Q(\theta)$. Now, I want to write the other roots as linear combinations of $1$, $\theta$, and $\theta^2$.. If we let $\alpha$ and $\beta$ be other roots, then $x^3-3x+1=(x-\theta)(x-\alpha)(x-\beta)$, from which we get that $\alpha+\beta=-\theta$ and $\alpha\beta=\theta^2-3$. Thus, $\alpha$ and $\beta$ are the roots of the quadratic $x^2+\theta x+(\theta^2-3)=0$. The discriminant of this quadratic is $12-3\theta^2$, so I'll be done if I can find rational $a, b$, and $c$ such that $(a+b\theta+c\theta^2)^2=12-3\theta^2$. But when I set up the equations, there was no obvious way to solve them, and even Wolfram Alpha ran out of computation time. Is there a less bashy way of doing this problem, or am I missing a trick in solving the resulting system of equations?
 A: $(a+b\theta+c\theta^2)^2 = 12-3\theta^2$ and $\theta^3 = 3\theta-1$ (from the initial polynomial, and the fact that plugging in $\theta$ will give zero).
Multiplying out gives
$\begin{align*}a^2+2ab\theta+(2ac+b^2)\theta^2+2bc\theta^3+c^2\theta^4 & = a^2+2ab\theta+(2ac+b^2)\theta^2+2bc(3\theta-1)+c^2(3\theta-1)\theta \\ & = a^2-2bc + (2ab+6bc-c^2)\theta+(2ac+b^2+3c^2)\theta^2\end{align*}$
From here, equate coefficients.  $a^2-2bc = 12, 2ab+6bc-c^2=0, 2ac+b^2+3c^2 = -3$
Wolframalpha comes up with $a=4,b=-1,c=-2$ or $a=-4,b=1,c=2$.  Checking these out, both solutions satisfy the system of equations.
A: If you compute the discriminant $\Delta$ of the cubic you will find that it's $81$, so it is a square. Therefore $(\theta-\alpha)(\theta-\beta)(\alpha-\beta) = \pm \sqrt \Delta = \pm 9$.
(if it's not a square, since the polynomial is irreducible, its Galois group is $S_3$ and not $A_3$, which implies that $\alpha$ and $\beta$ are not expressible in terms of $\theta$).
If you know $\theta$ then you know $\alpha + \beta = (\alpha + \beta + \theta) - \theta = - \theta$, and $\alpha\beta = \alpha\beta\theta / \theta = -1 / \theta$. So $\alpha - \beta = \pm 9 / (\theta-\alpha)(\theta-\beta) = \pm 9 / (\theta^2 + \theta - 1/\theta)$
Finally, $\alpha = \frac 12(\alpha + \beta + \alpha - \beta) = \frac 12(-\theta \pm 9/(\theta^2+\theta - 1/\theta))$
This procedure shows that there is a general formula showing that if $P = X^3+aX^2+bX+c$ and $\theta$ is a root of $P$ then $\alpha, \beta \in \Bbb Q(a,b,c,\theta,\sqrt {\Delta})$
