Hamiltonian cycle for 3-cube minus vertex

Question

If a vertex is removed from 3-cube, is it still Hamiltonian? I cannot find a Hamiltonian cycle, so if that's true how do I show that it has no Hamiltonian cycle?

Answer 1

The graph formed by removing a vertex from the cubical graph is not Hamiltonian, and this immediately follows from a 2-colouring of the vertices. There are four vertices of one colour and three of the other, but any Hamiltonian cycle must have an equal number of vertices of both colours, so no such cycle exists in the given graph.

Answer 2

How many vertices of degree $2$ does it have? What can you say about any Hamiltonian cycle with respect to the vertices it uses incident on a vertex of degree $2$?