Answer
$${\text{fifteenth term}} = 38760{x^6}{y^{42}}$$
Work Step by Step
$$\eqalign{
& {\text{fifteenth term of }}{\left( {x - {y^3}} \right)^{20}} \cr
& {\text{In the fifteenth term, }} - {y^3}{\text{ has an exponent of 15}} - 1,{\text{ or 14, }} \cr
& {\text{while }}x{\text{ has an exponent of 20}} - 14,{\text{ or 6}}{\text{.}} \cr
& \cr
& {\text{Using the }}k{\text{th Term of The Binomial Expansion}} \cr
& {\text{fifteenth term}} = \left( {20{\bf{C}}14} \right){\left( x \right)^6}{\left( { - {y^3}} \right)^{14}} \cr
& {\text{fifteenth term}} = \frac{{20!}}{{6!14!}}\left( {{x^6}} \right)\left( {{y^{42}}} \right) \cr
& {\text{fifteenth term}} = 38760{x^6}{y^{42}} \cr} $$
