<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] We know that $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$ 
Making the obvious substitutions, we get $$a^3+2b^3+4c^3-6abc=(a+b\sqrt[3]{2}+c\sqrt[3]{4})(a^2+b^2\sqrt[3]{4}+2c^2\sqrt[3]{2}-ab\sqrt[3]{2}-ac\sqrt[3]{4}-2bc),$$
so
$$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}.$$
For this specific problem, it is (more than) enough to find a polynomial $f(a,b,c)$ such that 
$$f(a,b,c)(a+b+c)=g(a^3,b^3,c^3),$$
for some other polynomial $g$ (this way the cube roots disappear). Although clever, the above solution I have above does not completely explain motivation. If we can rationalize the denominator of a fraction $1/q$, then it follows that $q$ must be algebraic. However, it is well known that the sum of algebraic numbers is itself algebraic (the proof itself is a bit more involved). Since all cube roots are (clearly) algebraic, it must be possible to rationalize the denominator. To rationalize the denominator of a fraction $1/q$, consider its minimal polynomial
$$Q(q)=\sum q_n q^n=0.$$
Note that
$$q\left(\dfrac{Q(q)-q_0}{q} \right)=-q_0,$$
which is clearly rational. Thus, in the example given in this problem, it is enough to find the minimal polynomial of $a+b\sqrt[3]{2}+c\sqrt[3]{4}$, which is simple yet painstaking.
In the case the above was not enough, we provide yet another way to get motivation. We want to multiply some factors to the denominator, currently an element of $\mathcal{S} = \mathbb{Q}[\sqrt[3]{2}] \setminus \mathbb{Q}$, to turn it into an element of $\mathbb{Q}$. Consider the map $F$ on $\mathbb{Q}[\sqrt[3]{2}]$ that sends
$$a + b\sqrt[3]{2} + c\sqrt[3]{4} \mapsto a + b\sqrt[]{2}\omega + c\sqrt[3]{4}\omega^2,$$
where $\omega = e^{2\pi i/3}$. The value of $\omega$ as a third root of unity is chosen so that $f$ is additive, multiplicative, and bijective.
The elements of $\mathbb{Q}$ are distinguished by being fixed by $F$. No element of $\mathcal{S}$ is fixed by $F$, so multiplying the given denominator by
$$(a + b\sqrt[3]{2}\omega + c\sqrt[3]{4}\omega^2)(a + b\sqrt[3]{2}\omega^2 + c\sqrt[3]{4}\omega),$$
we immediately produce an element fixed by $F$. It is easy to check that the resulting denominator and numerator are the same as we had earlier.
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