TITLE: Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?
QUESTION [35 upvotes]: We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. 
To be specific, it gives us 3 diophantine equations
$$a^{3}+2b^{3}+4c^{3}+12abc=3$$
$$a^{2}b+2b^{2}c+2c^{2}a=0$$
$$ab^{2}+2bc^{2}+ca^{2}=0$$
and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$. 
So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this. 
Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.

REPLY [6 votes]: Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $ \mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$.