'Quick' way to show non-regularity of languages that are 'close' to other non-regular langauges
It is known that regular languages are closed under homomorphisms. In particular, if $L$ were regular than so would $h(L) = \{b^nc^n : n \geq 0\}$ be, where $h$ is the homomorphism given by $h(a) = \epsilon$, $h(b) = b$, $h(c) = c$. Since $h(L)$ is not regular, so is $L$ non-regular.

More generally, you can use proof by contradiction together with operations under which regular languages are closed. The idea is to start with your language, perform a bunch of operations, and end up with a non-regular language. If all operations you used preserve regularity, then you can conclude that the original language is not regular.