Prove that $\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$ is greater than $\frac{2}{3}$. I have to prove that the product
$$\prod_{k=2}^{99}\frac{k^{3}-1}{k^{3}+1}$$
is greater than $\displaystyle\frac{2}{3}$.
I've tried to write $k^{3}-1$ as $(k-1)(k^{2}+k+1)$ or another ways but I couldn't finish it.

Use : $k^2 - k + 1 = (k-1)^2 + (k-1) + 1$ to cancel out some more terms.
