jmc

By Vieta's formulas, the sum of the three roots is $r+s+t=0$. Thus, we can write $$(r+s{)}^3+(s+t{)}^3+(t+r{)}^3=(-t{)}^3+(-r{)}^3+(-s{)}^3=-(r^3+s^3+t^3).$$Since each root satisfies the given equation, we have $$8r^3+1001r+2008=0,$$so $r^3=-\frac{1001}{8}r-251$. Similar equations hold for $s$ and $t$. Thus, $$-(r^3+s^3+t^3)=\frac{1001}{8}(r+s+t)+3\cdot 251.$$Since $r+s+t=0,$ the answer is $3\cdot 251=\boxed{753}$.