17. Let $z=\cos \frac{2 \pi}{2011}+i \sin \frac{2 \pi}{2011}$, and let
$$
P(x)=x^{2008}+3 x^{2007}+6 x^{2006}+\ldots \frac{2008 \cdot 2009}{2} x+\frac{2009 \cdot 2010}{2}
$$
for all complex numbers $x$. Evaluate $P(z) P\left(z^{2}\right) P\left(z^{3}\right) \ldots P\left(z^{2010}\right)$.
Answer: $2011^{2009} \cdot\left(1005^{2011}-1004^{2011}\right)$ Multiply $P(x)$ by $x-1$ to get
$$
P(x)(x-1)=x^{2009}+2 x^{2008}+\ldots+2009 x-\frac{2009 \cdot 2010}{2},
$$
or,
$$
P(x)(x-1)+2010 \cdot 1005=x^{2009}+2 x^{2008}+\ldots+2009 x+2010 \text {. }
$$

Multiplying by $x-1$ once again:
$$
\begin{aligned}
(x-1)\left(P(x)(x-1)+\frac{2010 \cdot 2011}{2}\right) & =x^{2010}+x^{2009}+\ldots+x-2010 \\
& =\left(x^{2010}+x^{2009}+\ldots+x+1\right)-2011 .
\end{aligned}
$$

Hence,
$$
P(x)=\frac{\frac{\left(x^{2010}+x^{2009}+\ldots+x+1\right)-2011}{x-1}-2011 \cdot 1005}{x-1}
$$

Note that $x^{2010}+x^{2009}+\ldots+x+1$ has $z, z^{2}, \ldots, z^{2010}$ as roots, so they vanish at those points. Plugging those 2010 powers of $z$ into the last equation, and multiplying them together, we obtain
$$
\prod_{i=1}^{2010} P\left(z^{i}\right)=\frac{(-2011) \cdot 1005 \cdot\left(x-\frac{1004}{1005}\right)}{(x-1)^{2}} \text {. }
$$

Note that $(x-z)\left(x-z^{2}\right) \ldots\left(x-z^{2010}\right)=x^{2010}+x^{2009}+\ldots+1$. Using this, the product turns out to be $2011^{2009} \cdot\left(1005^{2011}-1004^{2011}\right)$.