Find $\lim_{n \rightarrow\infty } P_n$whreas $P_n=\frac{2^3-1}{2^3+1}\cdot\frac{3^3-1}{3^3+1}\cdot\cdot\cdot\frac{n^3-1}{n^3+1}$.

There's some telescoping going on here. We have $$\frac{n^3-1}{n^3+1}=\frac{n-1}{n+1}\frac{n^2+n+1}{n^2-n+1}.$$ Then $$\prod_{n=2}^N\frac{n-1}{n+1}=\frac13\frac24\frac35\frac46\cdots\frac{N-1}{N+2}$$ and $$\prod_{n=2}^N\frac{n^2+n+1}{n^2-n+1}=\frac73\frac{13}7\frac{21}{13}\frac{31}{21}\cdots\frac{N^2+N+1}{N^2-N+1}.$$ Both these products telescope to give a closed form for $$\prod_{n=2}^N\frac{n^3-1}{n^3+1}.$$