The expression given is \( 8h^6 - 27j^9 \).

This expression is a difference of two cubes, which can be factored using the difference of cubes formula:

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

To apply this formula, you need to recognize each term as a cube:

1. \( 8h^6 = (2h^2)^3 \) because \( (2h^2)^3 = 8h^6 \).
2. \( 27j^9 = (3j^3)^3 \) because \( (3j^3)^3 = 27j^9 \).

Now, apply the difference of cubes formula:

\[
8h^6 - 27j^9 = (2h^2 - 3j^3)\left((2h^2)^2 + (2h^2)(3j^3) + (3j^3)^2\right)
\]

Let's expand and simplify the second factor:

- \((2h^2)^2 = 4h^4\)
- \( (2h^2)(3j^3) = 6h^2j^3 \)
- \( (3j^3)^2 = 9j^6 \)

Thus, the fully factored form is:

\[
8h^6 - 27j^9 = (2h^2 - 3j^3)(4h^4 + 6h^2j^3 + 9j^6)
\]

Would you like more details on this, or do you have any other questions?

### Related Questions:

1. What is the sum of cubes formula, and how does it differ from the difference of cubes?
2. How do you identify when an expression is a difference of cubes?
3. Can the expression \( 8h^6 - 27j^9 \) be factored further?
4. What are some real-world applications of factoring differences of cubes?
5. How can you factor the sum of cubes, and what is the formula?
6. What is the significance of recognizing perfect cubes in polynomials?
7. How do the coefficients affect the factoring process in cube-related expressions?
8. What happens if we add instead of subtract these two cubes?

#### Tip:
When dealing with polynomials, always check if the terms can be expressed as perfect cubes or squares. This can simplify the factoring process significantly.

Learn how to factor the expression 8h^6 - 27j^9 using the difference of cubes formula. Understand the step-by-step process with detailed explanations and examples. Suitable for advanced high school students and anyone interested in algebra and polynomial factoring.

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