Find the product of the series? Can anybody please help me to evaluate the following product?
$$\prod_{n=2}^\infty\dfrac{n^3 - 1}{n^3 + 1} $$

Notice $$\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$ and $(n+1)^2-(n+1)+1=n^2+n+1$. Therefore we have a telescoping product
$$\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}=\frac{1\cdot \color{blue}{7}}{\color{red}{3}\cdot 3}\cdot\frac{2\cdot \color{blue}{13}}{\color{red}{4}\cdot \color{blue}{7}}\cdot \frac{\color{red}{3}\cdot \color{blue}{21}}{\color{red}{5}\cdot \color{blue}{13}}\cdots=\frac{1\cdot 2}{3}=\frac{2}{3}$$
If nervous about doing cancelling like this in an infinite product (which is a valid point!), do it carefully for the partial products and end up with the same.
