We can use the Binomial Theorem to expand (2u−3v^3)^5 as follows:
\begin{align*} (2u - 3v^3)^5 &= \binom{5}{0}(2u)^5(-3v^3)^0 + \binom{5}{1}(2u)^4(-3v^3)^1 + \binom{5}{2}(2u)^3(-3v^3)^2 + \binom{5}{3}(2u)^2(-3v^3)^3 \ &\quad + \binom{5}{4}(2u)(-3v^3)^4 + \binom{5}{5}(-3v^3)^5 \ &= 32u^5 + 120u^4v^3 - 216u^3v^6 + 216u^2v^9 - 90uv^{12} + 27v^{15}. \end{align*}The coefficient of u^2v^9 is therefore −216.
