AMC12 2010 A

Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$. The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$. Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}$. We can visualize it has a $3$ by $3$ cube with a hollow $2$ by $2$ center, along with 6 "windows" cut out too. The cube formula gives the volume of the $3$ by $3$ cube as $3^3 = 27$ and $2^3 = 8.$ Additionally, the area of the "windows" is the 2x2 cut along with a thickness of $\frac{(3-2)}{2} = \frac{1}{2}.$ The volume of all 6 windows is $6(\frac{1}{2})(2)(2) = 12.$ This gives us our answer of $27 - 8 - 12 = \boxed{7\ \textbf{(A)}}.$ We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube. There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice. Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$. Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7\ \textbf{(A)}}$. We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners. Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box. $12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7\ \textbf{(A)}}$.