Denote $\lambda_c(G) $to be the smallest integer k (if it exists) such that there exists a set Y with k edges such that $G-Y$ is disconnected and each of its components has a cycle. let $G \notin \{K_4,K_{3,3}\}$ be a connected cubic graph such that $|Y|=\lambda_c(G)$ and $G-Y$ is disconnected and each component contains a cycle (there are exactly two)
Suppose the edges of G have been colored with three colors $1,2,3$ such that adjacent edges have diffrent colors and $n_i$ edges in Y have color $i=1,2,3$ prove that $n_1\equiv n_2\equiv n_3\equiv|Y| \mod 2 $
My best idea thus far has been to remove all the edges from the graph of color $n_1$ or otherwise stated consider the subgraph generated by colors $n_2$ and $n_3$ then look at the remaining cycles to try and determine the parity of $Y'$ i have also looked at removing all of the edges of one color from Y but i think this is less useful.
it may be useful to know that no edge in this graph is a bridge! consider if there was one consider the sub-graph generated by 2 colors including the color of the bridge its a union of cycles hence the edge lies on a cycle so it cannot be a bridge.
The first thing to notice is that in the sub-graph remaining that it is a chain of even cycles the next thing to notice is that $|Y'|\equiv0 \mod 2$ Notice in the picture no matter what side you are on of the split of Y' in order to make a cycle you must end on the same side as you started so the result follows. In fact we know that if we do the same method for any two of the 3 colors each new $|Y'|$ is $|Y'|\equiv0 \mod 2$.
It follows that we have $n_1+n_2 \equiv 0 \mod2$, $n_1+n_3 \equiv 0 \mod2$ and that $n_2+n_3 \equiv 0 \mod2$
it follows that $n_1\equiv n_2 \equiv n_3 \mod2$
it follows that $3n_1 \equiv |Y| \mod2 \implies n_1\equiv |Y| \mod 2$ Q.E.D