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.
 A: $x\sqrt[3]{2} = m - \sqrt[3]{2} \to \sqrt[3]{2} = \dfrac{m}{x+1} \to 2 = \dfrac{m^3}{x^3+3x^2+3x+1} \to 2x^3+6x^2+6x+2-m^3=0$. Thus $x$ is algebraic number.
A: Another proof: it is enough to prove $N(m-\sqrt[3]{2})$  is an integer. Now the norm is the product of the conjugates: if  $j=\mathrm e^{\tfrac{2\mathrm i\pi}3}$,
$$N\biggl(\frac{m-\sqrt[3]{2}}{\sqrt[3]{3}}\biggr)=\frac{(m-\sqrt[3]{2})(m-j\sqrt[3]{2})(m-j^2\sqrt[3]{2})}{\sqrt[3]3\cdot j\sqrt[3]{3}\cdot j^2\sqrt[3]{3} }\equiv \frac{m^3-2}3, $$
As $m\equiv 2\mod3$, $\,m^3\equiv 8\equiv 2\mod 3$, so $m^3-2\equiv 0\mod 3$, which proves the assertion.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            
A: 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!
A: Let $x=\frac{m-\sqrt[3]2}{\sqrt[3]3}$, and let $y=\sqrt[3]3x$.
Then $p(y)=(y-m)^3+2=0$.
Let $\omega$ be a complex cube-root of $1$.
$p(y)p(\omega y)p(\omega^2y)=q(y^3)=q(3x^3)$ is a cubic in $y$ with integer coefficients.  It is an integer polynomial in $x$ with lead coefficient $27$, and hopefully the other coefficients are multiples of $27$.
