已知 $x+y=3, x^{3}+y^{3}=36$, 则 $x y=$
解: $\because x+y=3$,

$\therefore(x+y)^{2}=9$,
则 $x^{2}+2 x y+y^{2}=9, \quad: x^{2}+y^{2}=9-2 x y$,

$\because x^{3}+y^{3}=(x+y) \quad\left(x^{2}+y^{2}-x y\right)=36$,

$\therefore 3(9-2 x y-x y)=36$,

则 $9-3 x y=12$,

故 $-3 x y=3$, 解得: $x y=-1$

此题主要考查了因式分解的应用, 关键是掌握 $x^{3}+y^{3}=(x+y)\left(x^{2}+y^{2}-x y\right)$.