<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] Magic answer: 
Let $f(n) =\dfrac{n(n-1)}{n^2-n+1}$. Then show $f(n+1) = \dfrac{n(n+1)}{n^2+n+1}$ and thus $$\frac{f(n)}{f(n+1)} = \frac{n(n-1)(n^2+n+1)}{n(n+1)(n^2-n+1)} = \frac{n^3-1}{n^3+1}$$
(I call this a "magic answer" just because most of the other answers here give you reasons for how you would see this, while I just plop in an $f$ that works, as if by magic. This is really the same argument as the others, just distilled to a minimalist essence.)
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