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Attaching a new leaf to a graph multiplies its chromatic polynomial by $t-1$, because after colouring the old graph the leaf may receive any colour except its neighbour's.

Factor the proposed polynomial:In particular,By the two-colourability criterion for bipartite graphs, every finite bipartite graph has at least one proper two-colouring, so its chromatic polynomial is positive at $2$. Therefore $f$ cannot be the chromatic polynomial of a bipartite graph.

It is nevertheless a chromatic polynomial. Start with the complete graph $K_3$, whose vertices may be coloured inways, and attach one leaf to any vertex. After the triangle is coloured, the leaf has $t-1$ available colours. The chromatic polynomial after attaching a leaf therefore gives

The inequality need not be strict. Start with the nearest-neighbour weighted graph ${\mathbb{Z}}^3$, attach a new leaf $b$ only to the origin $o$, and give its edge positive weight. Take $A=\{o\}$ and $B=\{o,b\}$. A walk starting from $b$ must move to $o$ at its first non-killed step, while any walk reaching $b$ from elsewhere must first pass through $o$. Hence the equilibrium potential of a finite set is the same for the two sets:Using $\mathrm{cap}(C)=\mathcal{E}(h_C,h_C)$ for a finite set $C$ givesdespite $A\ne B$.