Galois Theory and Galois Groups

I think it'd be way simpler to show that quadratic has no roots in $\,\Bbb Q(\sqrt[3]2)\,$ by analyzing its discriminant:

$$\Delta=\alpha^2-4\alpha^2=-3\alpha^2$$

But $\,\alpha=\sqrt[3]2\,$ , so $\,0>-3\alpha^2\in\Bbb R\,$ and from here the quadratic has no real roots, and this is enough as $\,K\,$ is a real field...

I'm of the opinion that you're working too hard.

Notice that $K$ is a real field so $\omega \sqrt[3]{2},\omega^2 \sqrt[3]{2} \notin K$ where $\omega=e^{2\pi i/3}$. Then the only root of $x^3-2$ in $K$ is $\sqrt[3]{2}$ so any automorphism fixes $\sqrt[3]{2}$.

Even though Don Antonio showed that the quadratic has no roots, I would like to point out that your three equations are easily seen to possess no solutions at all:
Firstly, since $a^2$ and $c^2$ are $\ge 0$, we find that both $c(1+2b)$ and $a(1+2b)$ are $\le 0$. But $1+b+b^2=(b+\frac{1}{2})^2+\frac{3}{4}$, thus $ac\le 0$. This implies that $a$ and $c$ are both of the same sign, and of the opposite sign, i.e. $a=c=0$, but then the third equation is not satisfied, so there is no solution at all, in real numbers.
Tell me if I miss something, thanks.