# proving that this family of angles, cannot be trisected

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?

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We prove an easy weaker result. Let $\alpha$ be a real transcendental. Then $x^3-\alpha$ is irreducible over $\mathbb{Q}(\alpha)$.
Suppose to the contrary that $x^3-\alpha$ is reducible over $\mathbb{Q}(\alpha)$. Then the real cube root $\alpha^{1/3}$ of $\alpha$ is in $\mathbb{Q}(\alpha)$. Thus there are polynomials $P$ and $Q$ with rational coefficients such that $$\alpha^{1/3}=\frac{P(\alpha)}{Q(\alpha)}.$$ Without loss of generality we may assume that the constant terms of $P$ and $Q$ are not both $0$. Take the cube of both sides. We obtain $\alpha Q^3(\alpha)=P^3(\alpha)$. This contradicts the fact that $\alpha$ is transcendental, since $P^3(x)-xQ^3(x)$ cannot be identically $0$.
To show that $P^3(x)-xQ^3(x)$ is not identically $0$, note this is clear if the constant term of $P$ is non-zero. If the constant term of $P$ is $0$, then the constant term of $Q$ is non-zero. But then the coefficient of $x$ in $xQ^3(x)$ is non-zero, while the coefficient of $x$ in $Q^3(x)$ is $0$.