(9) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{n}=(2 n-1)(2 n+1)(2 n+3)$, then the greatest common divisor $d$ of $a_{1}, a_{2}, \cdots$, $a_{2008}$ these 2008 terms is $\qquad$.
(9) On one hand, from $a_{1}=1 \times 3 \times 5, a_{2}=3 \times 5 \times 7, a_{3}=5 \times 7 \times 9$, $a_{4}=7 \times 9 \times 11$, we know $\left(a_{1}, a_{4}\right)=3$, so $d \mid 3$, thus $d \leqslant 3$; on the other hand, if $n \equiv 0(\bmod 3)$, then $2 n+3 \equiv 0(\bmod 3)$, thus $a_{n} \equiv 0(\bmod 3)$; if $n \equiv 1(\bmod 3)$, then $2 n+1 \equiv 0(\bmod 3)$, thus $a_{n} \equiv 0(\bmod 3)$; if $n \equiv 2(\bmod 3)$, then $2 n-1 \equiv 0(\bmod 3)$, thus $a_{n} \equiv 0(\bmod 3)$. Therefore, for any positive integer $n$, we have $3 \mid a_{n}$, so $3 \mid d$, and since $d \mid 3$, we have $d=3$.