TITLE: Is $\sqrt[3]{5}$ in $\mathbb Q(\sqrt[3]{2})$?
QUESTION [11 upvotes]: I am trying to solve question 3.7 (b) from Chapter 15 in Artin's book "Algebra".
The problem is:

Is it true that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$?

It is clear by Eisenstein's criterion that both $\sqrt[3]{2}$ and $\sqrt[3]{5}$ have their minimal polynomails of egree $3$ over $\mathbb Q$.
Thus if we assume that $\sqrt[3]{5}\in \mathbb Q(\sqrt[3]{2})$, then we must have $\mathbb Q(\sqrt[3]{5})=\mathbb Q(\sqrt[3]{2})$.
This didn't get me anywhere.
Then I tried the most simple minded approach.
Say $\sqrt[3]{5}=a+b\sqrt[3]{2}+c\sqrt[3]{4}$, for some $a,b,c\in \mathbb Q$.
By cubing and rearranging, it is likely that we will arrive at some absurdity.
But clearly this shouldn't be what the author must have intended.
Can somebody point me towards a slick solution?

REPLY [3 votes]: Here is an approach that is less elegant but more elementary than the answers so far: we examine the minimal polynomial of some algebraic combination of $\sqrt[3]{2}$ and $\sqrt[3]{5}$.
For example, $\sqrt[3]{2} + \sqrt[3]{5}$ satisfies the polynomial $x^9 - 21x^6 -123x^3  -343$, so we are done if we can show that this polynomial is irreducible.  This is not so easy to do by hand, of course.