Certainly! To factor the expression \(24x^3 - 81y^3\) completely, let's follow a systematic approach.
1. Recognize the special form: Notice that the expression \(24x^3 - 81y^3\) resembles a difference of cubes, which generally takes the form \(a^3 - b^3\).
2. Identify \(a\) and \(b\): To apply the difference of cubes formula, we need to write the given expression as a difference of cubes:
[tex]\[
24x^3 - 81y^3
\][/tex]
We can recognize this as:
[tex]\[
24x^3 - 81y^3 = (2x)^3 - (3y)^3
\][/tex]
Here, \(a = 2x\) and \(b = 3y\).
3. Apply the difference of cubes formula:
The difference of cubes can be factored using the formula:
[tex]\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
\][/tex]
Substituting \(a = 2x\) and \(b = 3y\):
[tex]\[
(2x)^3 - (3y)^3 = (2x - 3y)\left((2x)^2 + (2x)(3y) + (3y)^2\right)
\][/tex]
4. Simplify each part:
- \(a - b = 2x - 3y\)
- \(a^2 = (2x)^2 = 4x^2\)
- \(ab = (2x)(3y) = 6xy\)
- \(b^2 = (3y)^2 = 9y^2\)
Therefore, combining these:
[tex]\[
(2x - 3y)(4x^2 + 6xy + 9y^2)
\][/tex]
5. Finalize the factorization:
So, the completely factored form of the expression \(24x^3 - 81y^3\) is:
[tex]\[
24x^3 - 81y^3 = 3(2x - 3y)(4x^2 + 6xy + 9y^2)
\][/tex]
Thus, the fully factored form of \(24x^3 - 81y^3\) is:
[tex]\[
3(2x - 3y)(4x^2 + 6xy + 9y^2)
\][/tex]
