jmc

We have that $8=x^2+y^2=x^2+2xy+y^2-2xy=(x+y{)}^2-2xy=16-2xy$, therefore $xy=\frac{16-8}{2}=4$. Since $x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)(x^2+y^2-xy)$, we can directly substitute in the numerical values for each algebraic expression. This gives us $x^3+y^3=(4)(8-4)=\boxed{16}$.