10. [8] The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \ldots, r_{2007}$. A polynomial $P$ of degree 2007 has the property that $P\left(r_{j}+\frac{1}{r_{j}}\right)=0$ for $j=1, \ldots, 2007$. Determine the value of $P(1) / P(-1)$.
Answer: $\frac{289}{259}$. For some constant $k$, we have
$$
P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right) .
$$

Now writing $\omega^{3}=1$ with $\omega \neq 1$, we have $\omega^{2}+\omega=-1$. Then
$$
\begin{array}{c}
P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}{k \prod_{j=1}^{201}\left(-1-\left(r_{j}+\frac{1}{r_{j}}\right)\right)}=\prod_{j=1}^{2007} \frac{r_{j}^{2}-r_{j}+1}{r_{j}^{2}+r_{j}+1}=\prod_{j=1}^{2007} \frac{\left(-\omega-r_{j}\right)\left(-\omega^{2}-r_{j}\right)}{\left(\omega-r_{j}\right)\left(\omega^{2}-r_{j}\right)} \\
=\frac{f(-\omega) f\left(-\omega^{2}\right)}{f(\omega) f\left(\omega^{2}\right)}=\frac{\left(-\omega^{2007}+17 \omega^{2006}+1\right)\left(-\left(\omega^{2}\right)^{2007}+17\left(\omega^{2}\right)^{2006}+1\right)}{\left(\omega^{2007}+17 \omega^{2006}+1\right)\left(\left(\omega^{2}\right)^{2007}+17\left(\omega^{2} 2006+1\right)\right.}=\frac{\left(17 \omega^{2}\right)(17 \omega)}{\left(2+17 \omega^{2}\right)(2+17 \omega)} \\
=\frac{289}{4+34\left(\omega+\omega^{2}\right)+289}=\frac{289}{259} .
\end{array}
$$