<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] The following method may be practical for high school graduates:
Let $$y = x^2 + ax + b,\tag{1}$$ where $x = \sqrt[3]{z}$.
Then $$x^2 = y - ax - b.\tag{2}$$
Multiplying $(1)$ by $x$:
$$xy = z + ax^2 + bx.\tag{3}$$
Substituting $x^2$ from $(2)$ in $(3)$: $$xy = ay + (b - a^2)x - (ab - z). \tag{4}$$
Expressing $y$ from $(4)$: $$y = \frac{ux - v}{x - a},\tag{5}$$ where $u = b - a^2$ and $v = ab - z$.
Using the difference of cubes formula: $$(ux - v)(u^2x^2 + uvx + v^2) = u^3z - v^3.\tag{6}$$
Expressing $(ux - v)$ from $(6)$: $$ux - v = \frac{u^3z - v^3}{u^2x^2 + uvx + v^2}.\tag{7}$$
Substituting $(ux - v)$ from $(7)$in $(5)$: $$\frac{1}{y} = \frac{(x - a)(u^2x^2 + uvx + v^2)}{u^3z - v^3}.\tag{8}$$
Finally, opening parentheses in $(8)$: $$\frac{1}{y} = \frac{(uv - au^2)x^2 + (v^2 - auv)x + (u^2z - av^2)}{u^3z - v^3}.$$
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