$\underset{x\to \infty }{\lim }\frac{2^3-1^3}{2^3+1^3}\frac{3^3-1^3}{3^3+1^3}\dots ·\frac{n^3-1}{n^3+1}\text{ equals }$

$\begin{array}{l} \underset{n\to \infty }{\lim } \prod _{n=2}^n\frac{\left(n^3-1\right)}{\left(n^3+1\right)} = \underset{n\to \infty }{\lim } \prod _{n=2}^n \frac{\left(n-1\right)\left( n^2+n+1\right)}{\left(n+1\right) \left(n^2 -n+1\right)} \\ =\underset{n\to \infty }{\lim }\prod _{n=2}^n\frac{\left(n-1\right)}{n+1}\prod _{n=2}^n\frac{n^2+n+1}{n^2-n+1}=\\ = \underset{n\to \infty }{\lim }\frac{1·2·3·4·····\left(n-1\right)}{3·4·5·6·········\left(n-1\right)·n.\left(n+1\right)} \times \frac{7·13·21··········\left(n^2+n+1\right)}{3·7·13············\left(n^2-n+1\right)}\\ \\ =\underset{n\to \infty }{\lim }\frac{2}{n\left(n+1\right)}\times \frac{\left(n^2+n+1\right)}{3} =\underset{ }{\underset{n\to \infty }{\lim } \frac{2\left(1+{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n^2}}\right)}{3\left(1+{\displaystyle \frac{1}{n}}\right)} =\frac{2}{3}}\\ \\ \\ \\ \\ \end{array}$