# Hamiltonian cubic graphs contain at least three hamilton cycles.

I'm trying to prove using Smith's theorem (each edge on a cubic graph is contained in an even number of hamilton cycles) that if $G$ is $3-$regular and Hamiltonian then there exist at least three Hamilton cycles.
My idea was that since we already know the existence of a Hamiltonian cycle $C$, we can take an arbitrary vertex, $v$. For this $v$ we know that two out of three edges adjacent to it (say $e_1,e_2$) are being used for $C$. These two edges must be contained in at least one more hamilton cycle each, say $C_1, C_2$. If these two cycles do not coincide then we are done. But what about $C_1$ being the same with $C_2$? Then we could use the third edge to get another Hamilton cycle. So in order to complete the proof what i need is either an argument that allows me to say $C_1 \neq C_2$ (which i doubt) or an argument that makes sure that there exists a vertex $v \in V(G)$ such all three edges are contained in a hamilton cycle (and therefore in a non-zero even number of them).
So, assuming that all vertices have an edge adjacent that is not contained in any hamilton cycle i have trouble deducting a contradiction.
any hint on this would be appreciated.

We know that the graph contains a Hamiltonian cycle $H_1$. We choose an edge $e$ of $H_1$. We know from Smith's theorem that $e$ belongs to an even number of Hamiltonian cycles, therefore there exists a second Hamiltonian cycle $H_2$.
Now, we know that $E(H_1)\backslash E(H_2) \neq \emptyset$ (otherwise we would have that $H_1=H_2$). We also know that $e \not\in E(H_1)\backslash E(H_2)$. Therefore, there exists an edge $e'\neq e$ that belongs to $H_1$ but not $H_2$. From Smith's theorem again, we know that $e'$ belongs to an even number of Hamiltonian cycles, therefore there exists a third Hamiltonian cycle $H_3$.