To factor the expression $125 - b^6$, you can recognize it as a difference of two cubes. This requires rewriting $125$ and $b^6$ as cubes:

$125 = 5^3 \quad \text{and} \quad b^6 = (b^2)^3$

So the expression can be rewritten as:

$125 - b^6 = 5^3 - (b^2)^3$

Now, you can apply the difference of cubes formula:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Substituting $a = 5$ and $b = b^2$ into the formula gives:

$125 - b^6 = (5 - b^2)(5^2 + 5b^2 + (b^2)^2)$

Simplify further:

$125 - b^6 = (5 - b^2)(25 + 5b^2 + b^4)$

So, the factorization of $125 - b^6$ is:

$125 - b^6 = (5 - b^2)(25 + 5b^2 + b^4)$

Do you have any questions, or would you like further details?

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Related Questions:

1. How do you factor the sum of two cubes, $a^3 + b^3$?
2. Can you factor the expression $64 - a^6$?
3. What is the general formula for factoring a difference of cubes?
4. How can the difference of cubes be applied in solving polynomial equations?
5. Can you explain how to factor the expression $b^6 - 1$?
6. How would you factor $x^3 - 27$?
7. What is the relationship between the difference of cubes and synthetic division?
8. Can the factorization of a difference of cubes be extended to trigonometric expressions?

Tip: When working with higher powers, try rewriting the expression in a form that makes it easier to apply known formulas, like the difference or sum of cubes.