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?

$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$
We can use Euler's polyhedron formula that says that $F+V=E+2$.  We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more.  $6+8(3) = 30$.  In addition, each cube creates $7$ new vertices while taking away the original $8$, yielding $8(7) = 56$ vertices.  Thus $E+2=56+30$, so $E=\boxed{\textbf{(D) }84}$