I am attempting to solve a proof given in class which states the following:
A cubic tree is a tree whose vertices have degree either 1 or 3. Let T be a cubic tree and let G be a cubic graph obtained from T by adding a cycle through the leaves of T.
Show that G is 3-connected
The way I have attempted it to formulate my proof so far is supposing I have a cubic graph that was obtained from adding a cycle through the leaves of T. I also stated that Whitneys theorem tells us that that if the graph has 3 pairwise internally disjoint uv-paths then it must be 3 connected. So I begin attempting to show the graph has 3 pairwise disjoint uv-paths and this is where my logic gets lost. I started off by letting u,v by any two vertices in my cubic Graph G, and I also let P1,P2,P3 be three seperate paths in G. If I am somehow able to prove that each path is unique and only share u and v in common then I will have proved the result. So the problem lies in my way of going about showing each path is unique. Any hints?
If $u,v$ are both leaves of $T$, the desired property holds: There is one path within $T$ and two paths within the added cycle, obviously edge-disjoint.
For general $u,v$: First, we have the unique path $\gamma$ connecting $u$ and $v$ within $T$. If $u$ is not a leaf of $T$, we can start two paths from $u$ along its other two edges in $T$ and then continue these arbitrarily until a leaf. This gives two edge disjoint paths $\gamma_{u,1}$, $\gamma_{u,2}$ (also edge-disjoint to $\gamma$) from $u$ to some leaves $u'_1$, $u'_2$ of $T$. If $u$ is a already a leaf, then we can take $\gamma_{u,1}=\gamma_{u,2}=$empty path (and $u'_1=u'_2=u$). Do the same for $v$, resulting in $\gamma_{v,1}$, $\gamma_{v,2}$, $v_1'$, $v_2'$. In the added circle, we can find edge-disjoint paths either from $u_1'$ to $v_1'$ and $u_2'$ to $v_2'$ or from $u_1'$ to $v_2'$ and $u_2'$ to $v_1'$. Together with the $\gamma_{u,i}$, $\gamma_{v,i}$, these form the desired paths.