Prove that $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{-1}$ with a,b,c∈Q is a number of the form $d+e\sqrt[3]{2}+f\sqrt[3]{4}$ with $d,e,f∈Q$ Prove that  $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{-1}$ with $a,b,c∈Q$ is a number of the form $d+e\sqrt[3]{2}+f\sqrt[3]{4}$ with $d,e,f \in Q$
I'd like to do this without using too much fancy field-theoretic machinery. 
 A: Let $\omega =\sqrt[3]2$ then $$(a+b\omega+c\omega^2)(d+e\omega+f\omega^2)=(ad+2bf+2ce)+(ae+bd+2cf)\omega+(af+be+cd)\omega^2$$
The conditions for the product to be $1$ are $$ad+2bf+2ce=1$$
$$ae+bd+2cf=0$$
$$af+be+cd=0$$
Since $a,b,c$ are given these are three linear equations for the three unknowns $d,e,f$ which you can solve using your favourite method.
A: I see you included field theory, etc in the tags, so are you allowing at least a little bit of field-theoretic machinery?  If so, it's really not too terrible:
First, note that $\sqrt[3]{4} = (\sqrt[3]{2})^2$
Next, note that $\mathbb{Q}[\sqrt[3]{2}]$ is a degree $3$ field extension over $\mathbb{Q}$, since $f(x) = x^3 - 2$ is irreducible over $\mathbb{Q}$.  We can therefore take three linearly independent elements, say $1, \sqrt[3]{2}, (\sqrt[3]{2})^2 = \sqrt[3]{4}$ as a basis for this extension.
Using our basis, we know that any element of $\mathbb{Q}[\sqrt[3]{2}]$ can be written in the form $c_1 + c_2\sqrt[3]{2} + c_3\sqrt[3]{4}$ such that $c_i \in \mathbb{Q}$.  Since this is a field, then we are guaranteed that any number of this form has an inverse that can also be written in this form.
At any rate, I hope you're able to gain at least a little bit from my post, though Mark Bennet's answer is probably more along the lines of what you want.
