<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] Let $x=\frac {m-\sqrt [3] 2}{\sqrt [3] 3}$
Let $m=3k+2$
Then $x=\frac {3k+2-\sqrt [3] 2}{\sqrt [3] 3}$
$x\sqrt [3] 3=3k+2-\sqrt [3] 2$
$x\sqrt [3] 3+\sqrt [3] 2 =3k+2$
$\left( x\sqrt [3] 3+\sqrt [3] 2 \right)^3 =\left(3k+2 \right)^3$
$3x^3+3\sqrt [3] {18}x^2+3\sqrt [3] {12}x+2=27k^3+54k^2+36k+8$
... no not getting anywhere helpful!
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