To prove it, we can verify the equation easily for 

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[Not-so-daily puzzle] Infinite product

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The answer is $$2/3$$

The trick is to know the simplification:

$$
\frac{T^3-1}{T^3+1} = \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

This lets us prove by induction that:

$$
\prod_{n=2}^{T} \frac{n^3-1}{n^3+1} = \frac{2}{3} \left( \frac{T^2+T+1}{T^2+T} \right)
$$

which becomes $$2/3$$ in the limit as $$T \rightarrow \infty$$

To prove it, we can verify the equation easily for $$T=2$$.

Then:

$$
\prod_{n=2}^{T} \frac{n^3-1}{n^3+1} = \left( \prod_{n=2}^{T-1} \frac{n^3-1}{n^3+1} \right) \left( \frac{T^3-1}{T^3+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{(T-1)^2 + (T-1) + 1}{(T-1)^2 + (T-1)} \right) \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{T^2 - T + 1}{T^2-T} \right) \left( \frac{T-1}{T+1} \right) \left( \frac{T^2+T+1}{T^2-T+1} \right)
$$

$$
= \frac{2}{3} \left( \frac{T^2+T+1}{T^2+T} \right)
$$

Q.E.D.