Prove the ring $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ has inverse and is a field How can I prove that $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$ is of the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ (i.e. that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is a field)  for all rational $a,b,c$ and $a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0$?

This is very standard. Let $\phi(a,b,c)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. Then
you have
$$
\phi(a,b,c)+\phi(a',b',c')=\phi(a+a',b+b',c+c')
$$
$$
\phi(a,b,c)\phi(a',b',c')=\phi(aa'+2bc'+2cb',ab'+ba'+2cc',ac'+bb'+ca')
$$
$$
\phi(a,b,c)^{-1}=\phi(\frac{a^2-2bc}{D},\frac{2c^2-ab}{D},\frac{a^2-2bc}{D}), \
D=a^3+2b^3+4c^3-6abc
$$
Note that $a^3+2b^3+4c^3-6abc \neq 0$ whenever one of $a,b$ or $c$ is nonzero.
