S = {aⁿbcⁿ | n ≥ 0}.

Theorem. S is not regular.

Proof. By contradiction. Suppose there is a finite-state machine M that solves S. Run M on strings of the form aⁿ for n = 1, 2, 3, …. Since M has finitely many states, there must be two strings aⁱ and a^j, where i < j, that take M to the same state q.

From state q, continue with string bcⁱ, reaching state q'. Notice that aⁱbcⁱ and a^jbcⁱ must both take M to state q'. But the first of those strings is in S, while the second is not in S. Since q' cannot be both an accepting state and a rejecting state, that is a contradiction.