Answer: $(0, 0, 0)$ and $(t, t, t)$, where $t = \pm\sqrt{\frac{a+\sqrt{a^2+4bc}}{2c}}$.

Suppose that $x < y < z$ (one of the inequalities may not be strict). Then subtract the third equation from the first: $a(x^3 - y^3) + b(y - z) < 0 \le c(z^5 - x^5)$ – a contradiction. If we assume that $x < z < y$ (one of the inequalities may not be strict), then subtract the third equation from the second: $a(z^3 - y^3) + b(x - z) < 0 \le c(y^5 - x^5)$ – a contradiction. Without loss of generality, we can assume that all cases have been considered, since the system of equations is cyclic. Thus, we are left with the condition $x = y = z$.

To find $x$, we need to solve the equation: $cx^5 - ax^3 - bx = 0$. Obviously, $x_1 = 0$, then we need to solve the equation $cx^4 - ax^2 - b = 0$. Since $a^2 + 4bc > 0$ and
$$a - \sqrt{a^2 + 4bc} < 0, \text{ we have that } x_{2,3} = \pm\sqrt{\frac{a+\sqrt{a^2+4bc}}{2c}}, \text{ otherwise there are no other solutions.}$$