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

$$\dfrac{r^3-a^3}{r^3+a^3}=\dfrac{r-a}{r+a}\cdot\dfrac{r^2+ra+a^2}{r^2-ra+a^2}$$
Let $f(r)=r^2-ra+a^2$
$$f(r+h)=(r+h)^2-(r+h)a+a^2=r^2+r(2h-a)+a^2-ah$$
If we set $r^2+ra+a^2=f(r+h),$
$$r^2+ra+a^2=r^2+r(2h-a)+a^2-ah$$
Comparing the coefficients of $r,$ $$a=2h-a\iff h=a$$
Compare the constants $a^2=a^2-ah+h^2\iff h(h-a)=0$
$\implies h=a,$ $$r^2+ra+a^2=f(r+a)$$
$$\dfrac{r^3-a^3}{r^3+a^3}=\dfrac{f(r+a)}{f(r)}\cdot\dfrac{g(r-2a)}{g(r)}\text{ where } g(m)=m+a$$
Clearly 
$$\prod_{r=1}^n\dfrac{f(r+a)}{f(r)}\cdot\dfrac{g(r-2a)}{g(r)}$$ telescopes for finite integer $a$
