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.
 A: 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
A: 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, $$F=\frac{-29}{5}$$
A: I'll expand on my comments above.
You want to find a cubic polynomial satisfied by $y=x+x^{-1}$ whenever $x$ is a root of $P$.  Note that $x^3+3x+5=0$ gives $5x^{-1}=-(x^2+3)$ and $5x^{-3}=-(3x^{-2}+1)$.  Therefore
\begin{align*}
-5x^{-1}&=x^2+3\\
25x^{-2}&=(x^2+3)^2=x^4+6x^2+9\\
&=x(-3x-5)+6x^2+9\\
&=3x^2-5x+9\\
-125x^{-3} & = 25(3x^{-2}+1)\\
&=9x^2-15x+52.
\end{align*}
Thus
\begin{align*}
5y &=5x+5x^{-1}=-x^2+5x-3\\
25 (y^2-2) &=25 x^2 +25 x^{-2} =28x^2-5x+9\\
125 (y^3-3y) &=125 x^3 + 125 x^{-3} =-9x^2-360x-677
\end{align*}
Eliminating $x^2$ gives
\begin{align*}
25 (y^2-2)+140y &=135x-75\\
125 (y^3-3y)-45y &=-405x-650
\end{align*}
or equivalently,
\begin{align*}
5y^2+28y &=27x-5\\
25y^3-84y &=-81x-130
\end{align*}
Now eliminate $x$ and we have
\begin{align*}
-145=-81x-130+3(27x-5)&=(25y^3-84y)+3(5y^2+28y)\\
&=25y^3+15y^2
\end{align*}
i.e.
$$
5y^3+3y^2+29=0
$$
so the product of roots $y$ is $-29/5$.
Remark: This is essentially the "standard method" for finding a polynomial equation for $y_i=q(x_i)$ where $q$ is a polynomial and $x_i$ are the roots of some polynomial $P$.
