3. Find the value of the expression $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, if it is known that $a, b, c$ are three different real numbers satisfying the conditions:

$a^{3}-2022 a+1011=0, \quad b^{3}-2022 b+1011=0, \quad c^{3}-2022 c+1011=0$.
Answer: 2.

## Solution.

The cubic equation $t^{3}-2022 t+1011=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2022 t+1011: f(-100)0, f(10)0\right)$.

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

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

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