Baltic Way 1992 · Problem 9

Solution:

Consider the derivative $f'(x) = 3 x^{2} + 2 a x + b$. Since $b < 0$, it has two real roots $x_{1}$ and $x_{2}$. Since $f(x) \rightarrow \pm \infty$ as $x \rightarrow \pm \infty$, it is sufficient to check that $f(x_{1})$ and $f(x_{2})$ have different signs, i.e., $f(x_{1}) f(x_{2}) < 0$.

Dividing $f(x)$ by $f'(x)$ and using the equality $a b = 9 c$ we find that the remainder is equal to $x \left( \frac{2}{3} b - \frac{2}{9} a^{2} \right)$. Now, as $x_{1} x_{2} = \frac{b}{3} < 0$ we have $f(x_{1}) f(x_{2}) = x_{1} x_{2} \left( \frac{2}{3} b - \frac{2}{9} a^{2} \right)^{2} < 0$.