Every irreducible polynomial has at max. one root in $\mathbb{Q}(\sqrt[3]{2})$ 
Let $f \in \mathbb{Q}[x]$ be an irreducible polynomial. Show that $f$ has at maximum one root in $\mathbb{L} := \mathbb{Q}(\sqrt[3]{2})$.


My attempt:
If $f$ has no root in $\mathbb{L}$ then we're done. Assume $f$ has a root $\alpha_1$ in $\mathbb{L}$ that is not in $\mathbb{Q}$. Thus, it can be written as $f(x) = (x - \alpha_1)g(x) \in \mathbb{L}[x]$ for some $g(x) \in \mathbb{L}[x]$. Assume $g$ can also be written as $g(x) = (x - \alpha_2)h(x) \in \mathbb{L}[x]$ for a root $\alpha_2$ of $g$. Then we would have $f(x) = h(x)x^2 - h(x)(\alpha_1 + \alpha_2)x + h(x)\alpha_1\alpha_2$. Because $f \in \mathbb{Q}[x]$ we need $h(x) \in \mathbb{Q}[x]$ but also $h(x)(\alpha_1 + \alpha_2) \in \mathbb{Q}[x] \iff \alpha_1 + \alpha_2 \in \mathbb{Q}$ and $h(x)\alpha_1\alpha_2 \in \mathbb{Q}[x] \iff \alpha_1\alpha_2 \in \mathbb{Q}$.  Therefore, $\alpha_1$ and $\alpha_2$ must both be in $\mathbb{Q}$, contradicting our assumption that $\alpha_1 \notin \mathbb{Q}$. Therefore, $g$ has no roots in $\mathbb{L}$.
Though, I am not sure if the last part where I'm stating that $\alpha_1$ and $\alpha_2$ must be in $\mathbb{Q}$ is correct.

Another idea I had in my mind:
The only $\mathbb{Q}$-Automorphism of $\mathbb{Q}(\sqrt[3]{2})$ is the identity. Could we maybe argue that if $f$ had two roots in $\mathbb{L}$ then it wouldn't just be the identity?
 A: First, if $f$ has a root at all, then the degree of $f$ must be 1 or 3. This is because if $\alpha$ is a root of $f$, then we have $degree(f) = [\mathbb{Q}(\alpha) : \mathbb{Q}]$ is a factor of $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}]$, since we have a tower of field extensions $\mathbb{Q} \subseteq \mathbb{Q}(\alpha) \subseteq \mathbb{Q}(\sqrt[3]{2})$.
Now if $f$ is degree 1, then obviously we can’t have more than one root of $f$. So we need only concern ourselves with $f$ of degree 3. Suppose $f$ has multiple roots; then it must split completely in $\mathbb{Q}(\sqrt[3]{2})$. Furthermore, note that since both $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\alpha)$ are dimension 3 $\mathbb{Q}$-vector spaces, we can conclude the two are equal. Thus, $\mathbb{Q}(\sqrt[3]{2})$ is the splitting field of $f$. Therefore, the automorphism group of $\mathbb{Q}(\sqrt[3]{2})$ fixing $\mathbb{Q}$ is in fact a Galois group and thus has order $3 = [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}]$. But it can be easily shown that there is only one automorphism; contradiction.
A: Assume by contradiction that $f$ has at least 2 roots $\alpha,\beta\in\Bbb Q(\sqrt[3]2),$ then $2\le n:=\deg f=[\Bbb Q(\alpha):\Bbb Q]$ and $3=[\Bbb Q(\sqrt[3]2):\Bbb Q]=n[\Bbb Q(\sqrt[3]2):\Bbb Q(\alpha)],$ hence $n=3$ and the third root $\gamma$ of $f$ also belongs to $\Bbb Q(\sqrt[3]2)$ (since $\alpha+\beta+\gamma\in\Bbb Q$). Therefore, $\Bbb Q(\sqrt[3]2)=\Bbb Q(\alpha)$ is normal over $\Bbb Q,$ a contradiction.
A: If $f\in \mathbb{Q}[x]$ has a root $\alpha_1 \in \mathbb{Q}[\sqrt[3]{2}]$, then it also has roots $\alpha_2 \in \mathbb{Q}[\omega \sqrt[3]{2}]$, and $\alpha_3 \in \mathbb{Q}[\omega^2 \sqrt[3]{2}]$ ( $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$). Therefore, either $\alpha_1 \in \mathbb{Q}$, or $\alpha_j$'s are distinct and $(x-\alpha_1) (x-\alpha_2)(x-\alpha_3)$ is a rational factor of $f$.
