2. If $n$ is a natural number, and $n^{3}+2 n^{2}+9 n+8$ is the cube of some natural number, then $n=$ $\qquad$ .
2. 7 .

Since $n \in N$,
$$
\begin{aligned}
n^{3} & <n^{3}+2 n^{2}+9 n+8 \\
& <(n+2)^{3}=n^{3}+6 n^{2}+12 n+8,
\end{aligned}
$$

thus, only
$$
\begin{array}{l}
n^{3}+2 n^{2}+9 n+8 \\
=(n+1)^{3}=n^{3}+3 n^{2}+3 n+1 .
\end{array}
$$

Solving for the positive root $n=7$.