Solve the following system of equations in the set of positive numbers:

$$
\begin{aligned}
& x+y^{2}+z^{3}=3, \\
& y+z^{2}+x^{3}=3, \\
& z+x^{2}+y^{3}=3 .
\end{aligned}
$$
In the task, the roles of $x, y$, and $z$ are cyclically symmetric, meaning that by replacing $x$ with $y$, $y$ with $z$, and $z$ with $x$, the system of equations remains the same. Therefore, we can assume that among the three unknowns, $y$ is the (magnitude-wise) middle one, so either $x \leqq y \leqq z$ or $x \geqq y \geqq z$. The largest unknown cannot be less than 1, since the sum of the positive powers of three numbers less than 1 is less than 3. Similarly, it can be shown that the smallest unknown cannot be greater than 1. Therefore, in the two distinguishable cases, $x \leqq 1 \leqq z$, or $x \geqq 1 \geqq z$.

From the equality of the left sides of the first and second equations,

$$
y-1=\frac{1}{y}\left(\left(x^{3}-x\right)+\left(z^{2}-z^{3}\right)\right)
$$

follows, while from the first and third equations,

$$
y-1=\frac{1}{y^{2}}\left(\left(x-x^{2}\right)+\left(z^{3}-z\right)\right) .
$$

In the first case, the right side of (1) is the sum of two non-positive numbers, and the right side of (2) is the sum of two non-negative numbers; thus $y-1=0$, i.e., $y=1$. We get the same in the second case, since then the right side of (1) is non-negative, and the right side of (2) is non-positive. Substituting $y=1$ into the first two equations of the system:

$$
x+z^{3}=2=x^{3}+z^{2},
$$

thus

$$
x-x^{3}=z^{2}-z^{3}
$$

In the first case, therefore,

$$
0 \leqq x-x^{3}=z^{2}-z^{3} \leqq 0
$$

which means $x-x^{3}=z^{2}-z^{3}=0$, i.e., $x=1=z$. The only positive solution to the system of equations is therefore $x=y=z=1$.