7,8,9 |
| Calculate the product | $\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \ldots \cdot \frac{n^{3}-1}{n^{3}+1}$ |  |
Notice that $n^{2}-n+1=(n-1)^{2}+(n-1)+1$. Therefore,

$$
\begin{gathered}
\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \ldots \cdot \frac{n^{3}-1}{n^{3}+1}=\frac{(2-1)\left(2^{2}+2+1\right)}{(2+1)\left(2^{2}-2+1\right)} \cdot \frac{(3-1)\left(3^{2}+3+1\right)}{(3+1)\left(3^{2}-3+1\right)} \cdot \ldots \cdot \frac{(n-1)\left(n^{2}+n+1\right)}{(n+1)\left(n^{2}-n+1\right)}= \\
=\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \ldots \cdot \frac{n-1}{n+1} \cdot \frac{n^{2}+n+1}{2+1}=\frac{1 \cdot 2}{n(n+1)} \cdot \frac{n^{2}+n+1}{3}=\frac{2}{3}\left(1+\frac{1}{n(n+1)}\right)
\end{gathered}
$$

## Answer

$\frac{2}{3}\left(1+\frac{1}{n(n+1)}\right)$.