Factor 125, plus, b, cubed125+b 
3
  completely.

We are tasked with factoring the expression \(125 + b^3\).

We recognize that \(125\) is a perfect cube because:
\[
125 = 5^3
\]
So, the expression \(125 + b^3\) becomes:
\[
5^3 + b^3
\]

This matches the form of a **sum of cubes**, which can be factored using the identity:
\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]
In this case, \(a = 5\) and \(b = b\), so applying the formula:

\[
125 + b^3 = (5 + b)(25 - 5b + b^2)
\]

Thus, the fully factored form of \(125 + b^3\) is:
\[
(5 + b)(25 - 5b + b^2)
\]

Would you like any further explanation or details on this?

### Related Questions:
1. How would you factor a difference of cubes like \(b^3 - 125\)?
2. Can this factorization be applied to any sum of cubes?
3. What happens if \(b\) is negative in this sum of cubes?
4. How do we derive the identity for sum and difference of cubes?
5. Are there similar factorization identities for higher powers?

### Tip:
Always look for perfect squares or cubes in expressions—they often hint at special factorizations.

Factor 125, plus, b, cubed125+b^3 completely.

Learn how to factor the sum of cubes in the expression 125 + b^3 using the sum of cubes identity. This algebra problem is suitable for students in Grades 10-12 and involves recognizing perfect cubes and applying special factorization formulas.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation