Let $G$ be a $2$-connected $3$-regular graph. Is it true that $E(G) = E\_1 \cup E\_2$ where 
$G[E\_1]$(the induced subgraph on $E\_1$) is a cycle of $G$ and $G[E\_2]$ is a forest (Acyclic subgraph) of $G$?


**Edit:**


Since the counterexamples presented so far are $2$-connected, what happen if the connectivity of the graph is $3$ instead of two?



Is the set of edge of a cubic graph the union of a cycle and and an Acyclic graph?
Let $G$ be the cubic graph on $30$ vertices obtained by replacing each vertex of the [Petersen graph](http://en.wikipedia.org/wiki/Petersen_graph) with a triangle. Then $G$ is $3$-connected because the Petersen graph is $3$-connected, and there is no cycle in $G$ that meets all ten triangles because there is no Hamiltonian cycle in the Petersen graph.


