<s>[INST] How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$ 
How to compute
  $$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$

My Working : 
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
=  1-0 = 1$$
Is it correct
 [/INST] Note that
$${n^3 - 1 \over n^3 + 1} = {n - 1 \over n + 1}{n^2 + n + 1 \over n^2 - n + 1}$$
Also note that
$$(n-1)^2 + (n - 1) + 1 = n^2 - n + 1$$
So the infinite product in question is really the product of two telescoping products.
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