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 distinct real numbers satisfying the conditions:

$a^{3}-2020 a+1010=0, \quad b^{3}-2020 b+1010=0, c^{3}-2020 c+1010=0$.
Answer: 2.

Solution.

The cubic equation $t^{3}-2020 t+1010=0$ has three distinct roots (since for $\left.f(t)=t^{3}-2020 t+1010: f(-100)0, f(10)0\right)$.

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

$$
\left\{\begin{array}{l}
a+b+c=0 \\
a b+b c+a c=-2020 \\
a b c=-1010
\end{array}\right.
$$

We find the value of the expression: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+a c}{a b c}=\frac{-2020}{-1010}=2$.