Proving Irregularity of $L = \{ a^mb^nb^n \mid nm \ge 3 \} $
It is regular since $a^*(bb)^* \setminus L$ is regular:
$a^*(bb)^*\setminus L = a^* \cup (bb)^* \cup \{a^mb^{2n}|1\leq mn &lt; 3\}$
The last language of the union is finite thus regular.