4・137 Solve the system of equations
$$
\left\{\begin{array}{l}
x^{3}-y^{2}-y=\frac{1}{3}, \\
y^{3}-z^{2}-z=\frac{1}{3}, \\
z^{3}-x^{2}-x=\frac{1}{3} .
\end{array}\right.
$$
[Solution] The given equation is equivalent to the following system of equations:
$$
\left\{\begin{array}{l}
x=f(y), \\
y=f(z), \\
z=f(x) .
\end{array}\right.
$$

Here, \( f(t) = \sqrt[3]{t^2 + t + \frac{1}{3}} \). It is clear that for any real value, the function \( f(t) \) is always positive, so all solutions \((x, y, z)\) of the system of equations satisfy the inequalities \( x > 0, y > 0, z > 0 \). Moreover, when \( t > 0 \), the function \( f(t) \) is obviously monotonically increasing.

If \((x, y, z)\) is a solution of the system of equations and \( x > y \), then \( f(x) > f(y) \). From equation (1), we get \( z > x \), which implies \( z > y \) and \( f(z) > f(y) \). From equation (1) again, we get \( y > x \), leading to a contradiction. Similarly, if \( x < y \), a contradiction also arises. Therefore, \( x = y \), and similarly, we can prove \( y = z \).

Under the condition \( x = y = z \), the solution to the problem is \( x = y = z = \frac{1}{\sqrt[3]{4} - 1} \).