1. In the set of positive real numbers, solve the system of equations:

$$
\begin{aligned}
& x_{1}+x_{2}^{2}+x_{3}^{3}=3 \\
& x_{2}+x_{3}^{2}+x_{4}^{3}=3 \\
& x_{3}+x_{4}^{2}+x_{1}^{3}=3 \\
& x_{4}+x_{1}^{2}+x_{2}^{3}=3
\end{aligned}
$$
Solution. Let the quadruple $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ of positive real numbers be a solution to the given system of equations.

a) Suppose that at least two of the numbers $x_{1}, x_{2}, x_{3}, x_{4}$ are equal to 1. Since any two unknowns appear together in at least one equation, it is easy to see that the other two unknowns are also equal to 1.

b) Suppose that exactly one of the numbers $x_{1}, x_{2}, x_{3}, x_{4}$ is equal to 1. Let, for example, $x_{1}=1$. If $x_{2}<1$, then from the third equation, $x_{4}<1$, which is a contradiction. If $x_{2}>1$, then from the first, third, and fourth equations, we sequentially get $x_{3}<1$ and $x_{2}<1, x_{3}<1$. Then from the second equation, $x_{4}<1, x_{3}<1$. Let

$$
x_{1}=1-a, x_{2}=1+b, x_{3}=1-c, x_{4}=1+d, \text { where } 0<a<1, b>0, 0<c<1, d>0
$$

Then

$$
\begin{aligned}
0 & =x_{2}+x_{3}^{2}+x_{4}^{3}+x_{4}+x_{1}^{2}+x_{2}^{3}-x_{1}-x_{2}^{2}-x_{3}^{3}-x_{3}-x_{4}^{2}-x_{1}^{3} \\
& =\left(a^{3}-2 a^{2}+2 a\right)+\left(b^{3}+2 b^{2}+2 b\right)+\left(c^{3}-2 c^{2}+2 c\right)+\left(d^{3}+2 d^{2}+2 d\right)>0
\end{aligned}
$$

since $2 a>2 a^{2}$ and $2 c>2 c^{2}$, which is a contradiction.

v3) $x_{2}>1$. Then from the fourth equation, $x_{4}>1$, so

$$
3=x_{4}+x_{1}^{2}+x_{2}^{3}<x_{4}^{3}+x_{3}^{2}+x_{2}=x_{2}+x_{3}^{2}+x_{4}^{3}=3
$$

which is a contradiction.

v4) $x_{2}<1, x_{3}<1$. Then $x_{1}+x_{2}^{2}+x_{3}^{3}<3$, which is a contradiction.

Therefore, in the set of positive numbers, the only solution to the given system is $(1,1,1,1)$.