A $3 \times 3 \times 3$ cube has no Hamiltonian path starting at the corner.

We have a $3\times3\times 3$ cube which has $27$ cubes each $1\times1\times1$ stuck together as usual. $2$ cubes are neighbours if they have a common face. The corner cubes are the $8$ cubes at the corners of the big cube. Can the worm start from a corner cube and eat its neighbour cube, continuing until at the end the last cube eaten is the one in the center of the big cube which has 6 neighbour cubes?

Let me give a graphic theoretical argument

Hamiltonian Path: A path that visits each vertex exactly once

The problem statement is equivalent to determining whether there exists a Hamiltonian path on the 3-cube starting from a corner and ending at the center. We can describe the 3-cube as 27 lattice points. Without loss of generality, let the corner in which the mouse starts be the point $(0,0,0)$. Note that every edge along a path in the 3-cube will change the parity of the coordinate the mouse is at. A Hamiltonian path that traverses 27 vertices must contain exactly 26 edges.Therefore the origin and terminus of any Hamiltonian path on a 3-cube will be of the same parity. Since the center point is at $(1,1,1)$,there is no Hamiltonian path that starts at a corner and ends at the center. Done.