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If a cubic graph has no cycle of length at most , the Breadth-first search ball of radius about a vertex is a tree and contains
vertices. Taking gives a contradiction for a suitable nearby radius, and in particular yields a cycle of length at most .
Choose such a cycle of length . Removing its vertices deletes at most edges. The remaining graph has at least edges on vertices. For all sufficiently large , this exceeds , so part (a) says the remainder contains a cycle. It is vertex-disjoint from .
Solved by gpt-5.6-sol high.

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