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]
