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

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



Answer :

To factor the expression [tex]\( 8 x^{24} - 27 y^6 \)[/tex], we start by recognizing that it resembles the form of a difference of cubes, but raised to higher powers. Let's rewrite it to make this more apparent.

[tex]\[ 8 x^{24} - 27 y^6 \][/tex]

First, notice that [tex]\( 8 x^{24} \)[/tex] is equal to [tex]\((2x^8)^3\)[/tex] and [tex]\( 27 y^6 \)[/tex] is equal to [tex]\((3y^2)^3\)[/tex]. This lets us represent the expression in the form [tex]\( a^3 - b^3 \)[/tex], where [tex]\( a = 2x^8 \)[/tex] and [tex]\( b = 3y^2 \)[/tex]:

[tex]\[ (2x^8)^3 - (3y^2)^3 \][/tex]

The difference of cubes formula is:

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

Using this formula with [tex]\( a = 2x^8 \)[/tex] and [tex]\( b = 3y^2 \)[/tex], we get:

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

Now, let's expand the terms inside the second parenthesis:

[tex]\[ (2x^8)^2 = 4x^{16} \][/tex]

[tex]\[ (2x^8)(3y^2) = 6x^8 y^2 \][/tex]

[tex]\[ (3y^2)^2 = 9y^4 \][/tex]

Putting it all together, we get:

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

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

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

Thus, the correct choice is:

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