[Geometry]
Problem: Let $T=37$. A cube has edges of length $T$. Square holes of side length 1 are drilled from the center of each face of the cube through the cube's center and across to the opposite face; the edges of each hole are parallel to the edges of the cube. Compute the surface area of the resulting solid.

Solution: After the holes have been drilled, each face of the cube has area $T^{2}-1$. The three holes meet in a $1 \times 1 \times 1$ cube in the center, forming six holes in the shape of rectangular prisms whose bases are $1 \times 1$ squares and whose heights are $(T-1) / 2$. Each of these holes thus contributes $4(T-1) / 2=2(T-1)$ to the surface area, for a total of $12(T-1)$. Thus the total area is $6\left(T^{2}-1\right)+12(T-1)$, which can be factored as $6(T-1)(T+1+2)=6(T-1)(T+3)$. With $T=37$, the total surface area is $6(36)(40)=\mathbf{8 6 4 0}$.

Final answer: 8640