Irregularity of L = {a^i b^(j+3)| i!=j }
When you get stuck trying to prove a language is not regular by the Pumping Lemma, as in this example, because there doesn't seem to be a good choice of $u,v,w$, it's sometimes helpful to use closure properties first, to transform the problem. This is a good example of when this paradigm is useful.

We'll do this by contradiction. Suppose that $L$ were regular. Then since regular languages are closed under complement, $\overline{L}$ would be regular. Now $\overline{L}$ isn't particularly easy to describe, since it contains strings like $bababbba$, but what is $\overline{L}\cap a^*b^*$? This is just $a^nb^{n+3}=a^nb^nb^3$. Since regular languages are closed under intersection and $a^*b^*$ is regular, that would mean that $a^nb^nb^3$ would be regular. I'll bet you can now do a pumping lemma proof to show that this language wasn't regular.