8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4, n \in \mathbf{N}_{+}$, and let $d_{n}=\left(a_{n}, a_{n+1}\right)$, i.e., $d_{n}$ is the greatest common divisor of $a_{n}$ and $a_{n+1}$, then the maximum value of $d_{n}$ is $\qquad$.
8. 433 Detailed Explanation: From $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$ we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$, thus we have
$$
\begin{array}{l}
d_{n} \mid\left(-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right), 3 n^{2}+3 n+1\right), \\
d_{n}\left|\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right), \therefore d_{n}\right|\left(2 n+13,3 n^{2}+3 n+1\right), \text { hence } \\
d_{n} \mid(2 n+13,33 n-2) \text {, then } d_{n} \mid[-2(33 n-2)+33(2 n+13)], \text { i.e., } d_{n} \mid 433, \therefore\left(d_{n}\right)_{\max } \leqslant 433 .
\end{array}
$$

Let $n=210$, it is easy to see that $433 \mid a_{210}$ and $433 \mid a_{211}, 433=\left(a_{210}, a_{211}\right)$, thus $\left(d_{n}\right)_{\max }=433$.