hello, we have $ \prod_{k=2}^{100}\frac{k^3-1}{k^3+1}=\frac{3367}{5050}>\frac{2}{3}$ Sonnhard.

The product is telescopic in a way. Let $ f(n) = n - 1, g(n) = n^2 - n + 1$. Then- $ k^3 - 1 = f(k)g(k + 1), k^3 + 1 = f(k + 2)g(k)$. Thus, $ \prod_{k = 2}^{100}\frac {k^3 - 1}{k^3 + 1} = \prod_{k = 2}^{100}\frac {f(k)}{f(k + 2)}\frac {g(k + 1)}{g(k)} = \frac {f(2)f(3)}{f(101)f(102)}\frac {g(101)}{g(2)} =$ $ \frac {2}{10100}\frac {10101}{3} = \frac {10101}{3 \cdot 5050} = \frac {3367}{5050} > \frac {2}{3}$