3.13. $\left\{\begin{array}{l}\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}, \\ \left(x_{4}+x_{5}+x_{1}\right)^{5}=3 x_{2}, \\ \left(x_{5}+x_{1}+x_{2}\right)^{5}=3 x_{3}, \\ \left(x_{1}+x_{2}+x_{3}\right)^{5}=3 x_{4}, \\ \left(x_{2}+x_{3}+x_{4}\right)^{5}=3 x_{5} .\end{array}\right.$

![](https://cdn.mathpix.com/cropped/2024_05_21_3d5ef740640a71a48194g-033.jpg?height=402&width=443&top_left_y=787&top_left_x=1155)

## 3.3. Positive solutions
3.13. After cyclic renumbering of the unknowns, we can assume that $x_{1} \geqslant x_{i}(i=2,3,4,5)$. The function $f(x)=x^{5}$ is monotonically increasing, so $3 x_{2}=\left(x_{4}+x_{5}+x_{1}\right)^{5} \geqslant\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}$. Therefore, $x_{1}=x_{2}$ and $x_{3}=x_{1}$. Moreover, $3 x_{4}=\left(x_{1}+x_{2}+x_{3}\right)^{5} \geqslant$ $\geqslant\left(x_{5}+x_{1}+x_{2}\right)^{5}=3 x_{3}$. Therefore, $x_{4}=x_{3}$ and $x_{5}=x_{3}$.

We have obtained that $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=x$. This number $x$ must satisfy the equation $(3 x)^{5}=3 x$. We get three solutions: $x=0$ or $\pm 1 / 3$.