To what number $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}$ converges This question is from Ponnusamy and Silvermann's complex variables with applications Ch- 12 , subsection Infinite products.

Question: prove that $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}=2/3$.

I factored numerator and denominator to $\frac{(n-1) \times n^2+n+1 } { (n+1) \times n^2 -n+1}$ . But I think it's not possible to cancel denominantor and numerator in a way to get a pattren of cancelling due to factors $n^2-n+1$ and $n^2+n+1$ . So, I am struck on how should this be approached and unable to find any idea.
Kindly guide on how this should be approached.
Thanks!
Hint: The first five terms in the product $$\prod_{n = 2}^{\infty}\dfrac{n^3+1}{n^3-1} = \prod_{n = 2}^{\infty}\dfrac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)}$$ are $\dfrac{3 \cdot 3}{1 \cdot 7}$, $\dfrac{4 \cdot 7}{2 \cdot 13}$, $\dfrac{5 \cdot 13}{3 \cdot 21}$, $\dfrac{6 \cdot 21}{4 \cdot 31}$, and $\dfrac{7 \cdot 31}{5 \cdot 43}$. When you multiply these out, a lot of terms cancel. In general, when you multiply the first $N$ terms, a lot of stuff cancels, and you can simplify what's left. Then take the limit as $N \to \infty$.
Hint: The first five terms in the product $$\prod_{n = 2}^{\infty}\dfrac{n^3+1}{n^3-1} = \prod_{n = 2}^{\infty}\dfrac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)}$$ are $\dfrac{3 \cdot 3}{1 \cdot 7}$, $\dfrac{4 \cdot 7}{2 \cdot 13}$, $\dfrac{5 \cdot 13}{3 \cdot 21}$, $\dfrac{6 \cdot 21}{4 \cdot 31}$, and $\dfrac{7 \cdot 31}{5 \cdot 43}$. When you multiply these out, a lot of terms cancel. In general, when you multiply the first $N$ terms, a lot of stuff cancels, and you can simplify what's left. Then take the limit as $N \to \infty$.