It is known that every cubic Hamiltonian graph has at least three Hamiltonian cycles (by Tutte's theorem that every edge of a cubic graph belongs to an even number of Hamiltonian cycles)
Is it true that every cubic 3-connected planar Hamiltonian graph $G$ has three Hamiltonian cycles $C_1,C_2$ and $C_3$ such that every edge of $G$ belongs to exactly two of those cycles?
Clarification: we did not force that $G$ should have exactly three Hamiltonian cycles
Context: Quoting from Acyclically 4-colorable triangulations,
Observe that a triangulation $G$ ....(has property P)... if and only if its dual graph $G^*$ contains three Hamiltonian cycles such that each edge of $G^*$ is just contained in two of them. Since the problem of deciding whether a planar cubic 3-connected graph contains a Hamiltonian cycle is NP-complete[Garey,Johnson and Tarjan], we can deduce the problem of deciding whether a triangulation ....(has property P).... is NP-complete
(clarification in quote: triangulation - maximal planar graph [Not related to chordal completion] )
From their words, it sounds like the answer to the question is yes. But I am clueless as to why. I could not find any counter example either.
If there were constructions to make second (and third) Hamiltonian cycle from a given one, then the construction might be of use. But this is a hard problem in itself (even when the graph is guaranteed to have more than one Hamiltonian cycle).
It is easy to see that a cubic Hamiltonian graph has a 3-edge colouring. But, that doesn't seem to help either . (If we have a 3-edge colouring such that subgraph induced by edges coloured 1,2 is a Hamiltonian cycle, from that alone we can not say that edges coloured 2,3 (or 1,3) will form a Hamiltonian cycle. [I think most of the time they will not be]).