Có $(x^3+xy)^{15} = \sum_{k=0}^{15}.C^k_{15}.(x^3)^{15-k}.(xy)^k$$= \sum_{k=0}^{15}.C^k_{15}.x^{45-3k}.x^k.y^k$
$= \sum_{k=0}^{15}.C^k_{15}.x^{45-2k}.y^k$
$x^{21}.y^{12} \Rightarrow \left\{ \begin{array}{l} 45-2k=21\\ k=12 \end{array} \right. \Rightarrow k=12$
Hệ số chứa $x^{21}.y^{12}$ là $C^{12}_{15}=455$