Has $x^3 - x +1$ a root in $\mathbb{Q}(a)$?

Question

Decide if $x^3 - x +1$ has a root in $\mathbb{Q}(a)$, with $a \in\mathbb{C}$ such that $a^3 + a + 1 = 0$.

Answer 1

Based on the idea given by Jyrki Lahtonen, I think the following is relatively elementary.

Let $K$ be the splitting field of $x^3+x+1=0$ over $\mathbb{Q}$. Then the Galois group $G(K/\mathbb{Q})$ cannot fix the root of $x^3-x+1=0$ and therefore both cubics split in $K$.

However, this would mean that $K$ contained $L=\mathbb{Q}(\sqrt{-31},\sqrt{-23})$ and we would have $4=[L:\mathbb{Q}]$ as a factor of $6=[K:\mathbb{Q}]$.

Answer 2

We can write an arbitrary element of $\Bbb Q(a)$ as $x=\alpha+\beta a+\gamma a^2$, with $\alpha,\beta,\gamma\in\Bbb Q$. Let's plug that element into our equation: $$x^3-x+1=(\alpha+\beta a+\gamma a^2)^3 - (\alpha+\beta a+\gamma a^2) + 1.$$

You can expand this, reducing powers of $a$ by using the formula $a^3=-a-1$. Finally, you obtain something of the form $A+Ba+Ca^2$, where, $A$, $B$ and $C$ are expressions in $\alpha, \beta,\gamma$. Set all three to zero. If there is a rational solution, then you've shown that $x\in\Bbb Q(a)$