. Determine the functions $f$ from $\mathbb{Q}_{+}^{*}$ to $\mathbb{Q}_{+}^{*}$ satisfying $f(x+1)=f(x)+1$ and $f\left(x^{3}\right)=f(x)^{3}$ for all $x \in \mathbb{Q}_{+}^{*}$
. From $f(x+1)=f(x)+1$, we deduce $f(x+n)=f(x)+n$ for all natural integers $n$. Let $\frac{p}{q} \in \mathbb{Q}_{+}^{*}$ and let $u=f\left(\frac{p}{q}\right)$. We are given two expressions of $A=f\left(\left(\frac{p}{q}+q^{2}\right)^{3}\right)$. On one hand,

$$
A=f\left(\frac{p^{3}}{q^{3}}+3 p^{2}+3 p q^{3}+q^{6}\right)=u^{3}+3 p^{2}+3 p q^{3}+q^{6}
$$

since $f\left(\frac{p^{3}}{q^{3}}\right)=f\left(\frac{p}{q}\right)^{3}=u^{3}$ and $3 p^{2}+3 p q^{3}+q^{6}$ is a natural integer. On the other hand,

$$
A=f\left(\frac{p}{q}+q^{2}\right)^{3}=\left(u+q^{2}\right)^{3}=u^{3}+3 u^{2} q^{2}+3 u q^{4}+q^{6}
$$

By writing that the two expressions of $A$ are equal, we get

$$
3 u^{2} q^{2}+3 u q^{4}-\left(3 p^{2}+3 p q^{3}\right)=0
$$

This is a quadratic equation in $u$, of which $\frac{p}{q}$ is a root. Since the product of the roots is strictly negative, the other root is strictly negative, and we conclude that $u=p / q$. Thus, $f(x)=x$ for all $x \in \mathbb{Q}_{+}^{*}$.