Past exam of the Probability, statistics and options II course of the University of Oxford / 2008 / R / 2 / iv / a / Solution

Every cube vertex has degree three, so $Q_3$ is not Eulerian; it even has eight odd-degree vertices, so has no open Eulerian trail. After the four diagonals are added every vertex has degree four, and the graph remains connected. Thus $Q_3^{\prime }$ is Eulerian, by the circuit criterion proved above.