B
1) The best way to tackle this question is to think of the difference between two cubes:
![x^3-y^3=(x-y)(x^2+xy+y^2)](https://img.qammunity.org/2023/formulas/mathematics/college/w26p4ce81jywvwhexfkt2hdyggbr7c3ggb.png)
2) So now, let's apply to the binomial we have:
![\begin{gathered} \sqrt[3]{512}=8 \\ y^3-512=(y-8)(y^2+8y+64) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qr2gdn11dohbnxtvmoz58x5grktdz6lz2r.png)
So now, let's make use of the factor zero property for the first factor and solve the quadratic using the quadratic formula:
![\begin{gathered} y-8=0,y=8 \\ \\ y_=(-8\pm√(8^2-4\cdot\:1\cdot\:64))/(2) \\ y_1=(-8+8√(3)i)/(2)=\quad4+4√(3)i \\ y_2=(-8-8√(3)i)/(2)=\quad-4-4√(3)i \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k74vcyj7gqe93nceo8xp3sh0swujnt0rcq.png)
3) Thus, the answer is:
B