The expression $A^3 - 1$ can be factored using the difference of cubes formula. The formula for the difference of cubes is:

$x^3 - y^3 = (x - y)(x^2 + xy + y^2)$

In your case, $A^3 - 1$ can be written as:

$A^3 - 1^3$

Using the difference of cubes formula:

$A^3 - 1 = (A - 1)(A^2 + A \cdot 1 + 1^2)$

Simplifying further:

$A^3 - 1 = (A - 1)(A^2 + A + 1)$

So, the factored form of $A^3 - 1$ is:

$(A - 1)(A^2 + A + 1)$

Would you like more details or have any questions?

Here are some related questions to explore further:

1. How do you factor the sum of cubes, $A^3 + 1$?
2. What are the roots of the equation $A^3 - 1 = 0$?
3. How does the difference of cubes formula compare to the difference of squares?
4. Can $A^3 - 1$ be factored further if $A$ is a complex number?
5. What is the geometric interpretation of factoring $A^3 - 1$?

Tip: Remember that the difference of cubes formula can be applied to any terms raised to the third power, not just simple variables.