Three-edge path decomposition from a perfect matching

A finite simple graph that is a cubic graph and has a perfect matching admits a partition of a set of its edges into graph paths of length three.

Proof. Delete a perfect matching $M$. The remaining spanning subgraph has vertex degree two at every vertex, so it consists of disjoint graph cycles. Orient each cyclically, giving a successor bijection $s$. Replace each $uv\in M$ by the graph path $s(u),u,v,s(v)$. The four vertices are distinct: the successor edges are not in $M$, neither is a self-loop, and injectivity of $s$ excludes $s(u)=s(v)$. Every edge of $M$ appears once in the middle, and every other edge appears once at its unique tail. Thus the graph paths cover every edge exactly once. In particular, the perfect matching theorem for a cubic graph with at most two bridges guarantees this decomposition under its hypotheses.