From binomial expansion we have, that:
[tex](2x-y^2)=\sum\limits_{k=0}^9\binom{9}{k}\cdot(2x)^{9-k}\cdot(-y^2)^k=\\\\\\=
\dots+\binom{9}{4}\cdot(2x)^{9-4}\cdot(-y^2)^4+\dots=\\\\\\=
\dots+\binom{9}{4}\cdot2^5x^5\cdot(-1)^4(y^2)^4+\dots=
\dots+\binom{9}{4}\cdot2^5\cdot x^5y^8+\dots=\\\\\\=\dots+\dfrac{9!}{4!(9-4)!}\cdot32\cdot x^5y^8+\dots=\dots+\dfrac{9\cdot8\cdot7\cdot6}{4\cdot3\cdot2\cdot1}\cdot32\cdot x^5y^8+\dots=\\\\\\=\dots+126\cdot32\cdoy x^5y^8+\dots=\dots+\boxed{4032x^5y^8}+\dots[/tex]
Answer B.
