What is the factored form of [tex]\(8x^{24} - 27y^6\)[/tex]?

A. [tex]\((8x^8 - 27y^2)(2x^{16} + xy + 3y^4)\)[/tex]

B. [tex]\((2x^8 - 3y^2)(4x^{16} - 6x^8y^2 + 9y^4)\)[/tex]

C. [tex]\((2x^8 - 3y^2)(4x^{16} + 8x^8y^2 + 9y^4)\)[/tex]

D. [tex]\((8x^8 - 27y^2)(2x^{16} - 6xy + 3y^4)\)[/tex]



Answer :

To factor the given expression [tex]\( 8x^{24} - 27y^6 \)[/tex], we can identify the expression as a difference of cubes. The difference of cubes formula is:

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

First, we recognize the terms as cubes:

[tex]\[ 8x^{24} = (2x^8)^3 \][/tex]
[tex]\[ 27y^6 = (3y^2)^3 \][/tex]

Let’s identify [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

[tex]\[ a = 2x^8 \][/tex]
[tex]\[ b = 3y^2 \][/tex]

Now substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the difference of cubes formula:

[tex]\[ (2x^8)^3 - (3y^2)^3 = (2x^8 - 3y^2)\left((2x^8)^2 + (2x^8)(3y^2) + (3y^2)^2\right) \][/tex]

Simplify each term inside the parenthesis:

1. [tex]\((2x^8)^2 = 4x^{16}\)[/tex]
2. [tex]\( (2x^8)(3y^2) = 6x^8 y^2 \)[/tex]
3. [tex]\( (3y^2)^2 = 9y^4 \)[/tex]

Substituting back, we get the factored form:

[tex]\[ (2x^8 - 3y^2)\left(4x^{16} + 6x^8 y^2 + 9y^4\right) \][/tex]

Thus, among the given choices, the correct factored form is:

[tex]\[ \left(2x^8 - 3y^2\right)\left(4x^{16} + 6x^8 y^2 + 9y^4\right) \][/tex]

So, the correct answer is:

[tex]\[ \left(2 x^8 - 3 y^2 \right) \left(4x^{16} + 6x^8 y^2 + 9y^4\right) \][/tex]