Are there $x,y,z \in \mathbb Q \left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $x^2+y^2+z^2=-1$? The question in the headline was given as a practice exercise at the first quarter of an intro to Galois theory course.
I noticed that $\mathbb Q \left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ is an algebraic extension of degree $3$, since $\mathrm{irr} \left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3} \right)\right) = x^3-2$ (Eisenstein's criterion/general knowledge about complex roots of unit polynomials). 
Things got complicated from there. My intuition was that there aren't such $x,y,z$, but I was unable to prove that. I tried to assume that there are such $x,y,z$ and mess around with field extensions degrees, in order to arrive at a contradiction. This seemed promising, because we have
$$\mathbb{Q} \subset \mathbb{Q} \left(q\right) \subset \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3} \right)\right) \subset \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3} \right)\right) \left(q \right)$$
for $q=-(y^2+z^2+1)$ and $\left[\mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3} \right)\right): \ \mathbb{Q} \right]=3$. Thus, the multiplication theorem implies, among other things,
$$3\Big|\left[\mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3} \right)\right) (q): \ \mathbb{Q} \right].$$ However, whatever scheme I tried, I didn't manage to arrive at a contradiction. I then noticed that I can show that $\mathbb{Q}\left(\sqrt{q}\right)=\mathbb{Q}\left(q\right)$. That didn't lead me anywhere interesting, either.
I tried many different variations on these themes. Eventually, I got the sense that I wasn't able to arrive at a contradiction because none of my attempts really use the particular "structure" of the polynomial whose existence I was trying to disprove. Instead, my solutions depended on a too abstractly defined field extensions. That is, while I was playing with field extension of the form $K(q)$ or $K(\sqrt{q})$, the operations I tried on them didn't take into consideration the actual definition of $q$, which was derived from the given polynomial $x^2+y^2+z^2+1$. When I tried to develop this line of thought, I got stuck again. 
At this point, I feel I developed a mental block regarding this question. In my desperation, I even tried to prove the proposition, rather than disprove it. It didn't work. 
I'd love a nudge in the right direction. A hint, preferably not a subtle one. Thanks!
 A: Let $\omega = \exp\left(\frac{2\pi i}{3}\right)$.  As you noted $\Bbb{Q}[\omega\sqrt[3]{2}]$ is a root or stem field for $x^3-2$, so that we have an isomorphism $\sigma: \Bbb{Q}[\omega\sqrt[3]{2}]\mapsto \Bbb{Q}[\sqrt[3]{2}]$.  Suppose that there were $x,y$ and $z$ in $\Bbb{Q}[\omega\sqrt[3]{2}]$ such that $x^2+y^2+z^2=-1$, then we would have $\sigma(x)^2+\sigma(y)^2+\sigma(z)^2=-1$.  But this last equation is obviously impossible, since $\sigma(x), \sigma(y)$ and $\sigma(z)$ are real numbers.
A: This answer is based on @sharding4's and @JyrkiLahtonen's comments. I hope I formalized everything appropriately, give or take a few hand-waving explanations.

Let $g(x,y,z) = x^2+y^2+z^2+1$. 
$$E = \left\{e_1 = \sqrt[3]{2} \exp\left( \frac{2\pi i}{3}\right),\ e_2=e_1^2=\sqrt[3]{4} \exp\left( \frac{4\pi i}{3}\right), \ e_3=e_1^3 = 2 \right\}$$
is a basis of $\mathbb{Q}\left(\sqrt[3]{2} \exp\left( \frac{2\pi i}{3}\right)\right)$. Also,
$$F=\left\{f_1 = \sqrt[3]{2},\ f_2=f_1^2=\sqrt[3]{4}, \ f_3=f_1^3 = 2 \right\}$$
is a basis of $\mathbb{Q}\left(\sqrt[3]{2}\right)$.
Let $\varphi \left(\sum_{i=1}^3 q_i\cdot e_i \right) = \sum_{i=1}^3 q_i\cdot f_i$. It's easy to see that $$\varphi:\mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)\to\mathbb{Q}\left(\sqrt[3]{2}\right)$$ is an isomorphism of fields. Hence, there are $x,y,z\in \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $g(x,y,z)=0$ if and only if $$\varphi \left(g\left(x,y,z\right)\right)=g\left(\varphi \left(x\right),\varphi \left(y\right),\varphi \left(z\right)\right)=0.$$ But $g(a,b,c)\ge 1$ for all $a,b,c\in\mathbb{Q}\left(\sqrt[3]{2}\right)\subset\mathbb{R}$. Therefore, there are no $x,y,z \in \mathbb{Q}\left(\sqrt[3]{2} \exp\left(\frac{2\pi i}{3}\right)\right)$ such that $x^2+y^2+z^2=-1$.
