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.

Solutions Collecting From Web of "Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$"

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}
$

An overkill. Since we have $$\prod_{k\geq2}\frac{k^{3}-1}{k^{3}+1}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k^{2}+k+1\right)}{\left(k+1\right)\left(k^{2}-k+1\right)}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k-\frac{-1+i\sqrt{3}}{2}\right)\left(k-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+1\right)\left(k-\frac{1+i\sqrt{3}}{2}\right)\left(k-\frac{1-i\sqrt{3}}{2}\right)}$$ $$=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2-\frac{-1+i\sqrt{3}}{2}\right)\left(k+2-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+3\right)\left(k+2-\frac{1+i\sqrt{3}}{2}\right)\left(k+2-\frac{1-i\sqrt{3}}{2}\right)}
$$ and recalling the identity $$\prod_{k\geq0}\frac{\left(k+a_{1}\right)\left(k+a_{2}\right)\left(k+a_{3}\right)}{\left(k+a_{4}\right)\left(k+a_{5}\right)\left(k+a_{6}\right)}=\frac{\Gamma\left(a_{4}\right)\Gamma\left(a_{5}\right)\Gamma\left(a_{6}\right)}{\Gamma\left(a_{1}\right)\Gamma\left(a_{2}\right)\Gamma\left(a_{3}\right)},\, a_{1}+a_{2}+a_{3}=a_{4}+a_{5}+a_{6}
$$ which follows from the Euler’s definition of the Gamma function, we get $$P=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2-\frac{-1+i\sqrt{3}}{2}\right)\left(k+2-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+3\right)\left(k+2-\frac{1+i\sqrt{3}}{2}\right)\left(k+2-\frac{1-i\sqrt{3}}{2}\right)}=\frac{\Gamma\left(3\right)\Gamma\left(2-\frac{1+i\sqrt{3}}{2}\right)\Gamma\left(2-\frac{1-i\sqrt{3}}{2}\right)}{\Gamma\left(1\right)\Gamma\left(2-\frac{-1+i\sqrt{3}}{2}\right)\Gamma\left(2-\frac{-1-i\sqrt{3}}{2}\right)}
$$ and since $\Gamma\left(x+1\right)=x\Gamma\left(x\right)
$ we get $$P=\frac{2}{\left(2-\frac{1+i\sqrt{3}}{2}\right)\left(2-\frac{1-i\sqrt{3}}{2}\right)}=\color{red}{\frac{2}{3}}.$$