Looking for different approach to find $\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$ 
If $x_1,x_2,x_3$ be roots of $x^3+3x+5=0$, find the value of
$$\left(x_1+\frac1{x_1}\right)\left(x_2+\frac1{x_2}\right)\left(x_3+\frac1{x_3}\right)$$

To solve this problem I expanded the expression and used Vieta's formula and got the answer $-\frac{29}5$
. but I wonder is there any other approach to solve this problem?
I tried this way as an alternative ansewr:
Consider $x_1,x_2,x_3$ be roots polynomial $P(x)=x^3+3x+5$. then $\frac1{x_1},\frac1{x_2},\frac1{x_3}$ are roots of $P(\frac1x):$
$$P(\frac1x)=0\quad \rightarrow \quad P(\frac1x)=(\frac1x)^3+\frac3x+5=0\rightarrow 5x^3+3x^2+1=0$$
But I don't know whether it helps or not.

Another approach: Let the roots of $$x^3+3x+5=0 ~~~~(1)$$  be $a,b,c$, then $a+b+c=0, ab+bc+ca=3, abc=-5$.
Let $$F=(a+1/a)(b+1/b)(c+1/c)=\frac{(a^2+1)(b^2+1)(c^2+1)}{abc}=\frac{y_1 y_2y_3}{abc}=\frac{y_1 y_2 y_3}{-5}$$
where $y_1=(a^2+1)$ etc. let us transform $x$ equation to $y$ equation as $y=x^2+1 \implies x= \sqrt{y-1}.$ Putting it in (1) and simplifying we get $$y^3+3y^2-29=0$$ Hence $y_1y_2y_3=29$ and finally, $$F=\frac{-29}{5}$$
