jmc

We can rewrite $2007^3$ as the cube of a binomial: $$\begin{array}{rl}2007^3& =(2\cdot 10^3+7{)}^3\\ & =\left(\genfrac{}{}{0ex}{}{3}{0}\right)(2\cdot 10^3{)}^3+\cdots +(\genfrac{}{}{0ex}{}{3}{3})7^3\end{array}$$

Only the first and last term of the binomial expansion affect the values of $F$ and $L$. The first term is equal to $8\cdot 10^9$, so $F=8$. The last term is equal to 343, so $L=3$. Therefore, $F+L=\boxed{11}$.