ANSWER :
The answer is A.
EXPLANATION :
From the problem, we have :
[tex]\frac{4x^2-4x+1}{8x^3-1}[/tex]
The exclusion is when the denominator is equal to 0.
So equate the denominator to 0.
[tex]8x^3-1=0[/tex]
Factor completely using the difference of two cubes :
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
So that will be :
[tex]\begin{gathered} 8x^3-1=0 \\ (2x-1)(4x^2+2x+1)=0 \\ \text{ Equate both factors to 0} \\ 2x-1=0 \\ and \\ 4x^2+2x+1=0 \end{gathered}[/tex]
