2. Determine the values of $x \in Z$ for which $\sqrt[3]{x^{3}-6 x^{2}+12 x+29} \in Q$.
Solution: For the cube root of an integer to be rational, the number under the radical must be a perfect cube. (1p)

Therefore, there exists $y \in Z$ such that $x^{3}-6 x^{2}+12 x+29=y^{3} \Leftrightarrow(x-2)^{3}+$ $37=y^{3} \Leftrightarrow y^{3}-(x-2)^{3}=37 \Leftrightarrow(y-x+2)\left(y^{2}+y(x-2)+\right.$ $\left.(x-2)^{2}\right)=37$. (3p)

We obtain that $y-x+2$ is a divisor of 37, so $y-x+2=1$ and $y^{2}+$ $y(x-2)+(x-2)^{2}=37$, the latter expression being positive for any $\mathrm{x}$ and $y$. Solving the system, we get $x=-2$ and $y=-3$ or $x=5$ and $y=4$. (3p)