Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation $x^{3} + 3 x + 5 = 0$ . Then the value of expression $(x_{1} + \frac{1}{x_{1}}) (x_{2} + \frac{1}{x_{2}}) (x_{3} + \frac{1}{x_{3}})$ is equal to

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$- \frac{24}{5}$

$- 5$

$- \frac{29}{5}$

$- 1$

answer is A.

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Detailed Solution

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

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