6.346 For the equation $x^{3}+a x^{2}+b x+1=0$, express the product of the sum of its roots by the sum of their reciprocals in terms of the coefficients $a$ and $b$.
Solution. By Vieta's theorem $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-a, \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=b, \\ x_{1} \cdot x_{2} \cdot x_{3}=-1 .\end{array}\right.$

Therefore $\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-a \frac{x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}}{x_{1} x_{2} x_{3}}=a b$.

Answer: $a b$.