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.
Best Answer
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