Is $\mathbb{Q}[2^{1/3}]$ a field? Is $\mathbb{Q}[2^{1/3}]=\{a+b2^{1/3}+c2^{2/3};a,b,c \in \mathbb{Q}\}$ a field?
I have checked that $b2^{1/3}$ and $c2^{2/3}$ both have inverses, $\frac{2^{2/3}}{2b}$ and $\frac{2^{1/3}}{2c}$, respectively. 
There are some elements with $a,b,c \neq 0$ that have inverses, as $1+1*2^{1/3}+1*2^{2/3}$, whose inverse is $2^{1/3}-1$.
My problem is that is that I can't seem to find a formula for the inverse, but I also can't seem to find someone who doesn't have an inverse.
Thanks for your time.

Actually, the inverse of $x=2^{1/3}$ can be computed in a very elementary way as follows. Since $x\neq0$, the equality 
$$
x^3-2=x\left(x^2-\frac2x\right)=0
$$
yields
$$
\frac1x=\frac12x^2\in{\Bbb Q}[x].
$$
This "trick" extends immediately to any algebraic number.
