Help with recurrence solutions
Once you have $S(m) = 2 S(m-1) + 2^{3m}$, you can divide by $2^m$ and obtain :
$$\frac{S(m)}{2^m} = \frac{S(m-1)}{2^{m-1}} + 2^{2m}$$
Thus, setting $P(m) = \frac{S(m)}{2^m}$, you should be able to find $P(m)$, then $S(m)$.