62nd Ukrainian National Mathematical Olympiad

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