<s>[INST] Prove that $\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1}= \frac{3}{2}$, which is a beautiful value! Prove that $\prod\limits_{n= 2}^{\infty} \dfrac{n^{3}+ 1}{n^{3}- 1}= \dfrac{3}{2}$, which is a beautiful value ! A solution due to a p u as follow
$$\begin{align}
\prod\limits_{n= 2}^{\infty} \frac{n^{3}+ 1}{n^{3}- 1} & = \prod\limits_{n= 2}^{\infty}\frac{(n+ 1)(n+ \omega )(n+ \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\\ 
 & = \prod\limits_{n= 2}^{\infty} \frac{(n+ 1)(n- 1- \omega)(n- 1- \omega^{2})}{(n- 1)(n- \omega )(n- \omega^{2})}\,(\!1+ \omega+ \omega^{2}= 0\,used\!)\\ 
 & = \lim_{n\rightarrow \infty} \frac{n(n+ 1)(1- \omega)(1- \omega^{2})}{1\cdot 2(n- \omega)(n- \omega^{2})}\\ 
 & = \lim_{n\rightarrow \infty} \frac{3(1+ \frac{1}{n})}{2(1- \frac{\omega}{n})(1- \frac{\omega^{2}}{n})}\,\left (\!(1- \omega)(1- \omega^{2})= 1+ 1+ 1= 3\,used\!\right )\\ 
 & = \frac{3}{2}
\end{align}$$
I have a solution, and I'm looking forward to seeing a nicer one(s), thanks for your interests a lot !
 [/INST] Here is a much simpler way. Observe that the partial product
$$
\prod_{n=2}^{m} \frac{n^3+1}{n^3-1} = \frac{3 (m^2+m)}{2(m^2+m+1)}.
$$
Then when $m\to\infty$, the product converges to $\displaystyle \frac{3}{2}$.
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