3. Consider the function $f(x)=\sqrt[3]{1+x}+\sqrt[3]{3-x}$, defined on the real number line. For which $x$ is the value of the function an integer?
3. Answer: for $x \in\left\{1 \pm \sqrt{5}, 1 \pm \frac{10}{9} \sqrt{3}\right\}$.

Let $y=\sqrt[3]{1+x}, z=\sqrt[3]{3-x}$. Then $y^{3}+z^{3}=4$. Let $y+z=k \in \mathbb{Z}$. Clearly, $k \neq 0$. Then $\frac{4}{k}=\frac{y^{3}+z^{3}}{y+z}=y^{2}-y z+z^{2}=(y+z)^{2}-3 y z=k^{2}-3 y z$, from which $y z=\frac{k^{3}-4}{3 k}$.

The numbers $y, z$ turn out to be the roots of the quadratic equation $t^{2}-k t+\frac{k^{3}-4}{3 k}=0$. Its discriminant is $D=\frac{16-k^{3}}{3 k} \geq 0$. It is clear that $k$ cannot be negative, hence $0<k \leq \sqrt[3]{16}$, that is, $k=1$ or $k=2$.

In the first case, $y, z$ are the roots of the equation $t^{2}-t-1=0$, where $x=t^{3}-1$. Clearly, $t^{2}=t+1, t^{3}=t^{2}+t=2 t+1$, so $x=2 t=1 \pm \sqrt{5}$. It is quite obvious that this gives solutions to the original equation.

In the second case, $t^{2}-2 t+\frac{2}{3}=0$, from which $t^{2}=2 t-\frac{2}{3}, t^{3}=2 t^{2}-\frac{2}{3} t=\frac{10 t}{3}-\frac{4}{3}$. The roots have the form $t=1 \pm \frac{1}{\sqrt{3}}$, which gives $x=t^{3}-1=1 \pm \frac{10}{9} \sqrt{3}$. From general considerations, it is clear that the found values are suitable (since for them the system in $y$, $z$ gives a solution).

In total, there are four solutions in real numbers.