Answer
a. Eight smaller cubes (page 236, Figure 12.16).
b. Each face of the original cube has an area of $4 cm^{2}$. There are 6 faces, so the original total surface area is $24 cm^{2}$ .
Each of the eight smaller cubes has an area of $6 cm^{2} $, so the total surface area is $48 cm^{2}$. The ratio of new to old surface area is 2.
c. For the original cube: $\frac{24 cm^{2} }{8 cm^{3} } = 3 \frac{1}{cm}$.
For the set of 8 smaller cubes: $\frac{48 cm^{2} }{8 cm^{3} } = 6 \frac{1}{cm}$ , twice as great.
