4. Find the value of the expression $\frac{1}{a b}+\frac{1}{b c}+\frac{1}{a c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions $a^{3}-2022 a^{2}+1011=0, b^{3}-2022 b^{2}+1011=0, c^{3}-2022 c^{2}+1011=0$.
Answer: -2.

## Solution.

The cubic equation $t^{3}-2022 t^{2}+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t^{2}+1011: f(-3000)0, f(10)0\right)$.

Let these roots be $a, b, c$. Then, by Vieta's formulas:

$$
\left\{\begin{array}{l}
a+b+c=2022 \\
a b+b c+a c=0 \\
a b c=-1011
\end{array}\right.
$$

We find the value of the expression: $\quad \frac{1}{a b}+\frac{1}{b c}+\frac{1}{a c}=\frac{c+a+b}{a b c}=\frac{2022}{-1011}=-2$.