Prove that $\langle \{a+b\sqrt[3]2 \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not a field I got this question in my real analysis course:

Prove that $\langle \{a+b\sqrt{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is a field, but $\langle \{a+b\sqrt[3]{2} \mid a, b \in \mathbb{Q}\}, +, \cdot \rangle$ is not.

Obviously, I checked for the field properties (closure, commutativity, associativity, identity, inverse and distributivity) and they all check out except for the inverse. The additive inverse is trivial in both cases, and in the first case I found that the following is an inverse of $a+b\sqrt{2}$:
$$\frac{a}{a^2-2b^2}+\frac{-b}{a^2-2b^2}\sqrt{2}$$
$\forall$ $a, b \in \mathbb{Q}$. I have not been able to reproduce a similar result for the second set, which leads me to believe that a proof that the inverse does not always exist is the crux of the argument.
Any hints on how I can go about proving this?

Note: I have not had a course on abstract/modern algebra yet, so I only know the definition of a field and not any properties it may possess.

If $2^{2/3} = a+b2^{1/3}$ then $2^{1/3}$ is a root of $X^2-bX-a$ which means that $2^{1/3} = \frac{b\pm \sqrt{b^2+4a}}{2}$ an obvious contradiction.
(if you prefer then $X^3-2\in \Bbb{Q}[x]$ is irreducible so $2^{1/3}$ can't be the root of a degree two polynomial in $\Bbb{Q}[x]$)
