<s>[INST] "Rationalizing the denominator" of $1/(a + b\sqrt[3]{2} + c\sqrt[3]{4})$? If $(a, b, c) \in \mathbb{Q}^3 \setminus \{(0, 0, 0)\}$, so that $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ is a nonzero element of $\mathbb{Q}(\sqrt[3]{2})$, is there a formula for $${1\over{a + b\sqrt[3]{2} + c\sqrt[3]{4}}}$$ as a rational linear combination of $1$, $\sqrt[3]{2}$, and $\sqrt[3]{4}$?
 [/INST] Hint:
The product of conjugates elements is an integer (the norm of that element), which is equal to $2$ (see Viète's formulae). The norm of this element, setting $\omega=\mathrm e^{\tfrac{2\mathrm i\pi}3}$, is:
$$(a + b\sqrt[3]{2} + c\sqrt[3]{4})(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} + c\omega\sqrt[3]{4})=2.$$
Thus \begin{align*}(a + b\sqrt[3]{2} + c\sqrt[3]{4})^{-1}&=\frac12(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} + c\omega\sqrt[3]{4})\\
&=a^2-2bc+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}.\end{align*}
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