A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$. How many edges does the remaining solid have? {\small The solid after the eight corner cubes are removed.} $\textbf{(A) }36\qquad$ $\textbf{(B) }60\qquad$ $\textbf{(C) }72\qquad$ $\textbf{(D) }84\qquad$ $\textbf{(E) }108\qquad$
Solution
Removing a corner cube from each of the $8$ corners of the big cube replaces that vertex with $7$ new ones (all corners of the small cube except the shared one), so $V=8\cdot7=56$.
Each of the $12$ original edges survives (shortened) since both its endpoints are corners; each corner-removal also adds $9$ new edges from the exposed notch, giving $12+8\cdot9=84$ total edges.
A solid cube of side length $1$ is removed from four corners of a solid cube of side length $4$, no two of the four corners adjacent to each other. How many edges does the remaining solid have? {\small The solid after the four corner cubes are removed.}
