To factor the expression [tex]\(125a^6 - 64\)[/tex], let's begin by recognizing any patterns or special algebraic identities that might help. The expression [tex]\(125a^6\)[/tex] can be rewritten as [tex]\((5a^2)^3\)[/tex] and [tex]\(64\)[/tex] can be rewritten as [tex]\(4^3\)[/tex]. So, the expression can be seen in the form [tex]\( (5a^2)^3 - 4^3 \)[/tex].
This form [tex]\( x^3 - y^3 \)[/tex] suggests the use of the difference of cubes formula, which states:
[tex]\[
x^3 - y^3 = (x-y)(x^2 + xy + y^2)
\][/tex]
By setting [tex]\(x = 5a^2\)[/tex] and [tex]\(y = 4\)[/tex], we apply this formula:
[tex]\[
(5a^2)^3 - 4^3 = (5a^2 - 4)((5a^2)^2 + (5a^2)(4) + 4^2)
\][/tex]
Now, let's simplify the expression inside the parentheses:
1. [tex]\((5a^2)^2 = 25a^4\)[/tex]
2. [tex]\((5a^2)(4) = 20a^2\)[/tex]
3. [tex]\(4^2 = 16\)[/tex]
Putting it all together:
[tex]\[
(5a^2)^3 - 4^3 = (5a^2 - 4)(25a^4 + 20a^2 + 16)
\][/tex]
Hence, the factored form of [tex]\(125a^6 - 64\)[/tex] is:
[tex]\[
(5a^2 - 4)(25a^4 + 20a^2 + 16)
\][/tex]
Therefore, the correct option is:
[tex]\[
\boxed{(5a^2 - 4)(25a^4 + 20a^2 + 16)}
\][/tex]
