<s>[INST] 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?
 [/INST] $(a+b\theta+c\theta^2)^2 = 12-3\theta^2$ 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.
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