jmc

We have that $7=x^2+y^2=x^2-2xy+y^2+2xy=(x-y{)}^2+2xy=1+2xy$, therefore $xy=\frac{7-1}{2}=3$. 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 $x^3-y^3=(1)(7+3)=\boxed{10}$.