Compute $\prod\limits _{r=3 }^{\infty }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $ How should I go about evaluating this product? I have not been able to figure out.
$$\lim_{n\to\infty}\prod _{r=3 }^{n  }{  \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $$
 A: As an expansion of Did's comment, recognize the difference and the sum of cubes in the general factor to see the product can be rewritten as $$\prod_{r=3}^{\infty} \frac{(r-2)(r^2+2r+4)}{(r+2)(r^2-2r+4)}=   \prod_{r=3}^{\infty}\frac{r-2}{r+2}\frac{(r+1)^2+3}{(r-1)^2+3},$$ which makes clear that the $r$-th denominator of the first fraction and the $r+4$-th numerator of it, as well as  the $r$-th numerator of the second fraction and the $r+2$-th denominator of it will cancel each other, leaving $$\prod_{r=3}^\infty \frac{r^3-8}{r^3+8}= \frac{1\cdot2\cdot3\cdot4}{7\cdot12}=\frac{2}{7}.$$
A: We have 
$$\begin{align}
\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\prod_{n=3}^{N}\frac{(n-2)(n^2+2n+4)}{(n+2)(n^2-2n+4)}\\\\
&=\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}\tag 1
\end{align}$$

Now, analyzing the individual partial products we see that the first partial product reduces to 
$$\begin{align}
\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}&=\frac{1}{5}\frac{2}{6}\frac{3}{7}\frac{4}{8}\frac{5}{9}\cdots\frac{N-6}{N-2}\frac{N-5}{N-1}\frac{N-4}{N}\frac{N-3}{N+1}\frac{N-2}{N+2}\\\\
&=\frac{(1)(2)(3)(4)}{(N-1)(N)(N+1)(N+2)} \tag 2
\end{align}$$
while the second partial product reduces to 
$$\begin{align}
\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}&=\frac{4^2+3}{2^2+3}\frac{5^2+3}{3^2+3}\frac{6^2+3}{4^2+3}\cdots \frac{(N-1)^2+3}{(N-3)^2+3}\frac{(N)^2+3}{(N-2)^2+3}\frac{(N+1)^2+3}{(N-1)^2+3}\\\\
&=\frac{(N)^2+3}{2^2+3}\frac{(N+1)^2+3}{3^2+3}\tag 3
\end{align}$$

Putting $(2)$ and $(3)$ into $(1)$ yields
$$\begin{align}
\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\frac27 \frac{(N^2+3)((N+1)^2+3))}{(N+2)(N+1)(N)(N-1)}
\end{align}$$
from which we can easily see that
$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{\prod_{n=3}^{\infty}\frac{n^3-8}{n^3+8}=\lim_{N\to \infty}\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}=\frac27}
\end{align}$$
A: HINT: show by induction that $$\prod_{r=3}^k\frac{r^3-8}{r^3+8}=\frac{2(3+k^2)(4+2k+k^2)}{7(k-1)k(k+1)(2+k)}$$
