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.

It is a field, and you don't need to find an inverse for each element to prove that each element has an inverse. You can prove that if $\alpha$ is in the set then $\lbrace\,1,\alpha,\alpha^2,\alpha^3\,\rbrace$ is a linearly dependent set over the rationals, then deduce that $\alpha$ satisfies an equation of degree (at most) 3 over the rationals, then if $A\alpha^3+B\alpha^2+C\alpha+D=0$ you have $\alpha(A\alpha^2+B\alpha+C)=-D$, from which you can see an inverse to $\alpha$. 
