Consider the cube shown below with its 8 corners labelled a, b, c, d, e, f, g, and h. The figure is representative. All corners are to be coloured such that any two corners that are connected by an edge must be of different colours. The minimum number of colours required to achieve this is
Show Hint
A cube's corners split naturally into two groups of four so that every edge joins a corner from one group to a corner from the other; that alone is enough to colour it with just 2 colours.
Assign 3-bit binary coordinates to the 8 corners of the cube so that moving along any single edge changes exactly one coordinate. Colour every corner by the parity of the sum of its three coordinates: colour 1 if x+y+z is even, colour 2 if x+y+z is odd. Moving across any single edge flips exactly one coordinate, which always changes the parity of the sum by 1, so the two endpoints of every edge always land in different colour classes under this rule. This gives a valid proper colouring using only 2 colours.
Since the cube graph contains at least one edge, 1 colour can never work. As a 2-colouring was explicitly constructed above and 1 colour is impossible, the chromatic number, the minimum number of colours required, is exactly 2, confirming option (D) and ruling out 3, 4 and 8, all of which use more colours than necessary.
