30 Let $a$ and $b$ be the roots of $x^{2}+2000 x+1=0$ and let $c$ and $d$ be the roots of $x^{2}-2008 x+1=0$. Find the value of $(a+c)(b+c)(a-d)(b-d)$.
30. Answer: 32064
Note that $a b=c d=1, a+b=-2000$ and $c+d=2008$. So we have
$$
\begin{aligned}
(a+c)(b+c)(a-d)(b-d) & =\left[a b+(a+b) c+c^{2}\right]\left[a b-(a+b) d+d^{2}\right] \\
& =\left(1-2000 c+c^{2}\right)\left(1+2000 d+d^{2}\right) \\
& =(8 c)(4008 d) \\
& =32064
\end{aligned}
$$