We are asked to evaluate the product:

$\left(\frac{2^3 - 1}{2^3 + 1}\right) \times \left(\frac{3^3 - 1}{3^3 + 1}\right) \times \left(\frac{4^3 - 1}{4^3 + 1}\right) \times \cdots \times \left(\frac{20^3 - 1}{20^3 + 1}\right)$

Let's start by simplifying each individual fraction:

For any integer $n$, the expression can be written as:

$\frac{n^3 - 1}{n^3 + 1}$

This fraction can be factored as follows:

$\frac{n^3 - 1}{n^3 + 1} = \frac{(n - 1)(n^2 + n + 1)}{(n + 1)(n^2 - n + 1)}$

Notice that this expression has a telescoping pattern, where many terms will cancel out when multiplied together across the full product sequence.

Consider the product from $n = 2$ to $n = 20$:

$\prod_{n=2}^{20} \frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$

This telescoping product simplifies greatly. Most terms will cancel out except for the very first term of the first fraction and the very last term of the last fraction. Specifically, the simplified product reduces to:

$\frac{(2 - 1)(2^2 + 2 + 1)}{(20 + 1)(20^2 - 20 + 1)} = \frac{1 \times (4 + 2 + 1)}{21 \times (400 - 20 + 1)}$

Calculating this:

$\frac{1 \times 7}{21 \times 381} = \frac{7}{21 \times 381} = \frac{1}{3 \times 381} = \frac{1}{1143}$

Thus, the final answer is:

$\boxed{\frac{1}{1143}}$

Would you like to dive deeper into the simplification process or have any questions?

Here are five related questions for further exploration:

1. How does the telescoping property simplify complex product sequences?
2. What happens if the sequence started at $n = 1$ instead of $n = 2$?
3. How would the product change if the powers were different, e.g., $n^4$ instead of $n^3$?
4. Can you derive a general formula for such telescoping products?
5. What other mathematical sequences or series exhibit similar telescoping properties?

Tip: When dealing with large products or sums, always check if the expression can be simplified by telescoping, which can greatly reduce the complexity of the problem.