Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$ Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$
No idea how to even start. 

Using the suggested factorizations,
and using
$\begin{array}\\
k^2-k+1
&=k(k-1)+1\\
&=(k-1+1)(k-1)+1\\
&=(k-1)^2+(k-1)+1\\
\end{array}
$
(this is really the key),
$\begin{array}\\
\prod_{k=2}^n \dfrac{k^3-1}{k^3+1}
&=\prod_{k=2}^n \dfrac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)}\\
&=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n (k+1)}\dfrac{\prod_{k=2}^n (k^2+k+1)}{\prod_{k=2}^n (k^2-k+1)}\\
&=\dfrac{\prod_{k=2}^n (k-1)}{\prod_{k=2}^n (k+1)}\dfrac{\prod_{k=2}^n (k^2+k+1)}{\prod_{k=2}^n ((k-1)^2+(k-1)+1)}\\
&=\dfrac{\prod_{k=1}^{n-1} k}{\prod_{k=3}^{n+1}k}\dfrac{\prod_{k=2}^n (k^2+k+1)}{ \prod_{k=1}^{n-1} (k^2+k+1)}\\
&=\dfrac{2}{n(n+1)}\dfrac{n^2+n+1}{3}\\
&=\dfrac23\dfrac{ n^2+n+1}{n^2+n}\\
&=\dfrac23(1+\dfrac{ 1}{n^2+n})\\
& \to \dfrac23
\end{array}
$
