1. Given a quadratic trinomial $f(x)$ such that the equation $(f(x))^{3}-f(x)=0$ has exactly three solutions. Find the ordinate of the vertex of the trinomial $f(x)$.
# Answer: 0.

Solution. Suppose the leading coefficient of the trinomial is positive. Note that $(f(x))^{3}-f(x)=f(x) \cdot(f(x)-1) \cdot(f(x)+1)$. The equation $f(x)=0$ has more roots than the equation $f(x)=-1$, and fewer roots than the equation $f(x)=1$. It is also clear that no two equations have common roots. Therefore, the equation $f(x)=0$ has exactly one root. Consequently, the ordinate of the vertex of the trinomial $f(x)$ is zero. Similarly, the case where the leading coefficient of the trinomial is negative is also considered.