<s>[INST] Evaluate $P(1)/P(-1)$ 
The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.

Let $P(x) = a_{2007}x^{2007} + ... + a_0$. 
$r_j + \frac{1}{r_j} = \frac{r_j^2 + 1}{r_j}$
So we need the map: $x + \frac{1}{x} \to x$. 
We need a $u \to x$. Where $u(x + 1/x) = x$. 
The inverse function is: $u = \frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4})$ [USED WOLFRAM ALPHA]
So,
$P(x) = \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2007} + 17 \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 - 4}) \bigg)^{2006} + 1$
But letting $x= 1$ or $-1$ makes it so that the answer is imaginary!
 [/INST] Continuing after step $(2)$ in @Jack D'Aurizio's answer.
$$\begin{align}\frac{P(1)}{P(-1)}&=\prod_{i=1}^{2007}\frac{r_i^2-r_i+1}{r_i^2+r_i+1}=\prod_{i=1}^{2007}\frac{(r_i+\omega)(r_i+\omega^2)}{(r_i-\omega)(r_i-\omega^2)}=\prod_{i=1}^{2007}\frac{(-\omega-r_i)(-\omega^2-r_i)}{(\omega-r_i)(\omega^2-r_i)}\\ &=\prod_{i=1}^{2007}\frac{f(-\omega)\ f(-\omega^2)}{f(\omega)\ f(\omega^2)}\\ &=\frac{(17\omega^2)(17\omega)}{(2+17\omega^2)(2+17\omega)}=\frac{289}{259}\end{align}$$
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