10. If the sequence $\left\{a_{n}\right\}$ has the sum of the first $n$ terms
$$
\begin{array}{l}
S_{n}=n^{3}-n^{2}\left(n \in \mathbf{Z}_{+}\right), \\
\text { then } \sum_{i=1}^{2015} \frac{1}{a_{i}+8 i-2}=
\end{array}
$$
10. $\frac{2015}{6048}$.

Notice,
$$
a_{n}=\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n-1} a_{i}=3 n^{2}-5 n+2 \text {. }
$$

Also, $a_{1}=0$, then
$$
\begin{array}{l}
a_{n}=3 n^{2}-5 n+2\left(n \in \mathbf{Z}_{+}\right) . \\
\text {Therefore, } \sum_{i=1}^{2015} \frac{1}{a_{i}+8 i-2}=\sum_{i=1}^{2015} \frac{1}{3 i(i+1)} \\
=\frac{1}{3} \sum_{i=1}^{2015}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\frac{2015}{6048} .
\end{array}
$$