Question

$\left( x_{1} + \frac{1}{x_{1}} \right)\left( x_{2} + \frac{1}{x_{2}} \right)\left( x_{3} + \frac{1}{x_{3}} \right)$

$$\begin{matrix} = x_{1}x_{2}x_{3} + \frac{x_{1}x_{2}}{x_{3}} + \frac{x_{2}x_{3}}{x_{1}} + \frac{x_{1}x_{3}}{x_{2}} + \frac{x_{1}}{x_{2}x_{3}} + \frac{x_{2}}{x_{3}x_{1}} + \frac{x_{3}}{x_{1}x_{2}} + \frac{1}{x_{1}x_{2}x_{3}} \\ = x_{1}x_{2}x_{3} + \frac{x_{1}^{2}x_{2}^{2} + x_{2}^{2}x_{3}^{2} + x_{1}^{2}x_{3}^{2}}{x_{1}x_{2}x_{3}} + \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}{x_{1}x_{2}x_{3}} + \frac{1}{x_{1}x_{2}x_{3}} \\ x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = \left( x_{1} + x_{2} + x_{3} \right)^{2} - 2\left( x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} \right) = - 2(3) = - 6 \\ x_{1}^{2}x_{2}^{2} + x_{2}^{2}x_{3}^{2} + x_{3}^{2}x_{1}^{2} = \left( x_{1}x_{2} + x_{2}x_{3} + x_{3}x_{1} \right)^{2} - 2x_{1}x_{2}x_{3}\left( x_{1} + x_{2} + x_{3} \right) = 9 \\ \Rightarrow \ \left( x_{1} + \frac{1}{x_{1}} \right)\left( x_{2} + \frac{1}{x_{2}} \right)\left( x_{3} + \frac{1}{x_{3}} \right) = - 5 + \frac{9}{- 5} + \frac{- 6}{- 5} + \frac{1}{- 5} = - \frac{29}{5} \end{matrix}$$