64j3 - 27k3 = ?

Click on your answer.

(4j+3k)(16j2-12jk+9k2)
(4j-3k)(16j2-12jk+9k2)
(4j+3k)(16j2+12jk-9k2)
(4j-3k)(16j2+12jk+9k2)

The given expression is \( 64j^3 - 27k^3 \).

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

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

Here, \( a = 4j \) and \( b = 3k \), so:

\[
64j^3 - 27k^3 = (4j - 3k)((4j)^2 + (4j)(3k) + (3k)^2)
\]

Expanding the terms inside the second factor:

\[
= (4j - 3k)(16j^2 + 12jk + 9k^2)
\]

Thus, the correct answer is:

**(4j - 3k)(16j^2 + 12jk + 9k^2)**

Would you like details or have any questions?

Here are 8 follow-up questions for further practice:
1. How would you factor \( 125a^3 + 343b^3 \)?
2. What is the general method for factoring any difference of cubes?
3. Factor \( 8x^3 - 27y^3 \).
4. What happens if you factor \( 8m^3 + 125n^3 \)?
5. Can you explain why \( a^3 - b^3 \) and \( a^3 + b^3 \) have different factoring patterns?
6. Expand \( (5p - 2q)^3 \) using binomial expansion.
7. Solve the equation \( 27y^3 - 64z^3 = 0 \).
8. What is the role of the middle term when factoring a sum or difference of cubes?

**Tip:** When factoring, double-check that each term of the polynomial fits the pattern you're using to avoid mistakes.

Learn how to factorize the expression 64j^3 - 27k^3 using the difference of cubes formula. This problem involves algebraic manipulation and application of key formulas in polynomial factorization. Suitable for high school students studying algebra.

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