125 - b⁶ using sum and difference of two cubes

To express \( 125 - b^6 \) using the sum and difference of two cubes, we can proceed as follows:

### Step 1: Recognize the expressions as cubes
Notice that:
- \( 125 = 5^3 \)
- \( b^6 = (b^2)^3 \)

So the expression \( 125 - b^6 \) can be rewritten as:
\[
125 - b^6 = 5^3 - (b^2)^3
\]

### Step 2: Apply the difference of cubes formula
The difference of cubes formula is given by:
\[
a^3 - c^3 = (a - c)(a^2 + ac + c^2)
\]
Here, \( a = 5 \) and \( c = b^2 \). Substituting these into the formula gives:
\[
5^3 - (b^2)^3 = (5 - b^2)\left(5^2 + 5b^2 + (b^2)^2\right)
\]

### Step 3: Simplify the expression
Simplify the terms in the second factor:
\[
5^2 = 25
\]
\[
(b^2)^2 = b^4
\]
So the expression becomes:
\[
5^3 - (b^2)^3 = (5 - b^2)(25 + 5b^2 + b^4)
\]

### Final Answer
The expression \( 125 - b^6 \) can be factored as:
\[
125 - b^6 = (5 - b^2)(25 + 5b^2 + b^4)
\]

Would you like further details or have any questions?

Here are some related questions you might find interesting:

1. How do you factor \( a^3 + b^3 \) using the sum of cubes formula?
2. Can you show an example of factoring \( x^3 - 8 \)?
3. What is the general formula for factoring the sum of two cubes?
4. How can the difference of squares be applied in more complex factorizations?
5. What is the relationship between cubes and quadratic forms?
6. How do you simplify higher power binomials using cube formulas?
7. How do cube roots relate to the factorization of polynomials?
8. Can you explain how the factorization process differs when dealing with sum versus difference of cubes?

**Tip:** When factoring expressions involving powers, always check if you can express the terms as powers of smaller numbers, like squares or cubes, to apply known formulas.

Learn how to factor 125 - b^6 using the sum and difference of cubes formula. This problem involves algebraic manipulation with cubes and applies the difference of cubes formula to simplify the expression. Suitable for students in Grades 9-12.

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