LP relaxation of independent set
Non-bipartite connected cubic graphs have the unique optimal solution $x_i = 1/2$; in a bipartite cubic graph you have an integral optimal solution.



Proof: In a cubic graph, if you sum over all $3n/2$ constraints $x_i + x_j \le 1$, you have $\sum_i 3 x_i \le 3n/2$, and hence the optimum is at most $n/2$.

The solution $x_i = 1/2$ for all $i$ is trivially feasible, and hence also optimal.

In a bipartite cubic graph, each part has half of the nodes, and the solution $x_i = 1$ in one part is hence also optimal.

Any optimal solution must be tight, that is, we must have $\sum_i 3 x_i = 3n/2$ and hence $x_i + x_j = 1$ for each edge $\{i,j\}$. Thus if you have an odd cycle, you must choose $x_i = 1/2$ for each node in the cycle. And then if the graph is connected, this choice gets propagated everywhere.