Given a field $ \Bbb Q \subset K \subset \Bbb C$. One can prove that $\beta \in \Bbb C$ is constructible over $K$ iff the galois group of the minimal polynomial over $K$, $m_{\beta}(x)\in K[x]$ is a 2-group.
Thus given $\alpha \in \Bbb C$ with $ |\alpha|=1$, one want to prove if it's possible to construct $\alpha^{\frac{1}{3}}$ in other words if it's possible to trisect the angle given by $\alpha$. Then using the above, one has to consider the polynomial $x^3-\alpha \in Q(\alpha)=K$, clearly the minimal polynomial of $\alpha$ divides $x^3-\alpha$ and the galois group of a cubic could be $ S_3$ or $\Bbb Z_3$ none of them is a 2-group. So the criteria is the following:
The important of all this
Given $\alpha \in \Bbb C$ with $|\alpha|=1$ it's $\underline{not}$ possible to trisect the angle given by $\alpha$ (construct $\alpha^{\frac{1}{3}}$) if $x^3-\alpha \in \Bbb Q(\alpha)$ irreducible.
The problem:
Given $\alpha \in \Bbb C$ such that $|\alpha|=1$ and trascendental. Then it's not possible trisect $\alpha$. To prove this using the lemma it's just enough to prove that $x^3-\alpha$ is irreducible over $\Bbb Q(\alpha)$.
If $\beta$ is a root, then $\beta^3 = \alpha$ the other roots are given by $\beta , \beta \zeta_3 , \beta \zeta_3^2$. How supposing that it's reducible I can I prove that this contradicts the fact that $\alpha$ is trascendental?