<s>[INST] Show that $\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ is an algebraic integer. Let $m$ be an integer such that $m \equiv 2 \pmod 3$. Show that the number $$\dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$$ is an algebraic integer.
The usual technique, doing $x = \dfrac{m - \sqrt[3]{2}}{\sqrt[3]{3}}$ and trying to find an algebraic expression in terms of $x$ seems not to work in this case (at least I couldn't do it). Can anyone help me? This is a question from an old exam.
 [/INST] $x\sqrt[3]{2} = m - \sqrt[3]{2} \to \sqrt[3]{2} = \dfrac{m}{x+1} \to 2 = \dfrac{m^3}{x^3+3x^2+3x+1} \to 2x^3+6x^2+6x+2-m^3=0$. Thus $x$ is algebraic number.
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