To factorize $a^3 - 1$ in the form involving $(a-1)$ or $(1-a)$, we can apply the difference of cubes formula:

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

So, the factorization of $a^3 - 1$ is:

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

To express it in the form involving $(1-a)$, we can factor out $-1$ from $(a - 1)$, giving:

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

Thus, the two forms are:

1. $(a - 1)(a^2 + a + 1)$
2. $-(1 - a)(a^2 + a + 1)$

Would you like more details on how the factorization works or have any questions about this process?

Related Questions:

1. How do you factorize other types of polynomials like $a^3 + 1$?
2. What is the general formula for factoring higher powers like $a^n - 1$?
3. How does the difference of cubes formula apply in real-world problems?
4. What happens if we factor $a^3 - 1$ over complex numbers?
5. Can you prove why $a^2 + a + 1$ is part of the factorization?

Tip: Always remember to check if you can factor out common terms first before applying special formulas like the difference of cubes.