8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$
8.433.

Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(2 n+13,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid(2 n+13,33 n-2) \\
\Rightarrow d_{n} \mid[-2(33 n-2)+33(2 n+13)] \\
\Rightarrow d_{n} \mid 433 .
\end{array}
$$

Therefore, $\left(d_{n}\right)_{\max } \leqslant 433$.
It is easy to see that, $\left(a_{210}, a_{211}\right)=433$.
Thus, $\left(d_{n}\right)_{\max }=433$.