Evaluating $\prod_{n=2}^\infty {n^3-1\over n^3+1}$ 
The value of the infinite product $$P = \frac 79 \times \frac{26}{28} \times \frac{63}{65} \times \cdots \times \frac{n^3-1}{n^3+1} \times \cdots$$ is
(A) $1$
(B) $2/3$
(C) $7/3$
(D) none of the above

I wrote first 6 terms and tried to cancel out but did not get any idea what will be the last term
I did one same kind of problem $\prod (1-{1\over k^2}),k\ge2$ whose answer is $1\over 2$
here I  think the answer will be $D$, none of this?

HINT:
Observe that $$n^3-1=(n-1)(n^2+n+1)\text{ and }n^3+1=(n+1)(n^2-n+1)$$
Again, $\displaystyle (n+1)^2-(n+1)+1=\cdots=n^2+n+1$
Just set a few values of $n$ to find the surviving terms 
