jmc

If we cube both sides of the first equation, we find that $x^3+3x^{2}y+3x{y}^2+y^3=729$, so $x^3+y^3=729-(3x^{2}y+3x{y}^2)$. Since $3x^{2}y+3x{y}^2=3(xy)(x+y)=3(10)(9)$, we see that $x^3+y^3=729-(3x^{2}y+3x{y}^2)=729-270=\boxed{459}$.