Generalization. Let $G = (V,E)$ be a finite simple graph on $|V|=n$ vertices, and let $\delta(G) = \min\{\deg v \mid v\in V\} \geq 3$ be the minimum degree of $G$. Then for any positive integer $\lambda > 2$ the graph $G$ contains a cycle $\gamma$ of length not divisible by $\lambda$ (but that is not necessarily true for $\lambda = 2$). (In our case, with $\Delta(G) = \max\{\deg v \mid v\in V\}$, we have $\delta(G) = \Delta(G)$, i.e. the graph is $3$-regular, or cubic, and $\lambda = 3$.) Let $P = v_0v_1\ldots v_k$ be a longest path contained in $G$. This means all neighbours of $v_0$ are contained in $P$, so there must exist $1<i<j\leq k$ so that $v_0v_i, v_0v_j \in E$. Then $\gamma_1 = v_0v_1\ldots v_iv_0$ is a cycle of length $i+1$, $\gamma_2 = v_0v_1\ldots v_jv_0$ is a cycle of length $j+1$, and $\gamma_3 = v_0v_i\ldots v_jv_0$ is a cycle of length $j-i+2$. Assume $\lambda$ divides all three lengths; then $\lambda \mid (j+1) - (i+i) = (j-i+2)-2$ forces $\lambda \mid 2$, absurd. But for $\lambda = 2$ and the cube graph $ABCDA'B'C'D'$, the graph is bipartite, of shores $\{A, B', C, D'\}$ and $\{A', B,C',D\}$, therefore all its cycles are of even length.