Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$

we have to find $$\frac{(x_1^2+1)(x_2^2+1)(x_3^2+1)}{x_1x_2x_3}$$ now $$P(x)=x^3+3x+5=(x-x_1)(x-x_2)(x-x_3)$$ Notice that $$P(i)P(-i)=(x_1^2+1)(x_2^2+1)(x_3^2+1)$$ now use $x_1x_2x_3=-5$ to finish..

$i=\sqrt{-1}$ and $P(i)P(-i)$ can be easily found and is left as an exercise