question involving limit of product

Question

I got stuck with this problem and it is unlike any problem I've encountered till now .
Let $$P_n=\frac{2^3-1}{2^3+1}. \frac{3^3-1}{3^3+1} .\frac{4^3-1}{4^3+1}.....\frac{n^3-1}{n^3+1}$$
we have to prove that $\lim_{n\to\infty}P_n=\frac{2}{3}$.
I could solve this type of problem if instead of product of terms I'd been given sum, by finding general term and all.
But, I'm not able to think much after I saw the product. Maybe we can use sandwich theorem . but, i dont know how? Please help! Thanks in advance.

Answer 1

Hint:

$n^3-1=(n-1)(n^2+n+1)$

$n^3+1=(n+1)(n^2-n+1)$

$(n+1)^2-(n+1)+1=n^2+n+1$

Use this. The product telescopes.