Factor h, cubed, minus, 64, w, cubedh 
3
 −64w 
3
  completely.

To factor \( h^3 - 64w^3 \) completely, we recognize this as a difference of cubes. The formula for factoring a difference of cubes is:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\]

Here, \( a = h \) and \( b = 4w \) (since \( 64w^3 = (4w)^3 \)).

Using the formula, we have:

\[
h^3 - 64w^3 = (h - 4w)(h^2 + 4wh + (4w)^2)
\]

Simplify further:

\[
h^3 - 64w^3 = (h - 4w)(h^2 + 4wh + 16w^2)
\]

Thus, the complete factorization of \( h^3 - 64w^3 \) is:

\[
(h - 4w)(h^2 + 4wh + 16w^2)
\]

Would you like further details on any steps?

Here are some related questions to explore:
1. How do you factor the sum of cubes?
2. Can you solve \( h^3 + 64w^3 \) using a similar method?
3. What are the steps for factoring a perfect square trinomial?
4. How do you recognize the difference between a difference of cubes and a difference of squares?
5. Can you apply this method to more complex polynomials?

**Tip:** Always check if a polynomial fits special patterns like difference or sum of cubes or squares before factoring.

Learn how to factor h^3 - 64w^3 using the difference of cubes formula. This problem is ideal for students in Grades 9-11 and covers algebraic concepts including factoring and recognizing special patterns in polynomials.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation