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^ Clearly we only have to consider the numbers $\{1, 2, 3, 4, 6, 7, 9\}.$ As $\{1, 8\}$ are already cubes we can consider them later. Now we have the set $\{2, 3, 4, 6, 9\}.$ From that we have $\{2, 4\}, \{3, 9\}, \{2, 4, 3, 9\},$ and $\{4, 6, 9\}.$ There are $2^2 = 4$ ways to take subsets from $\{1, 8\}$ and $4$ sets not containing $1$ and $8$. So in total we have $4 \cdot 4 + 3 = \boxed{19}$ because there are $4 \cdot 4$ ways to create subsets and there is also $2^2 - 1$ ways to create subsets from $\{1, 8 \}$ (empty set does not count).