\begin{align*} L = \left \{ a^nb^{n+k} \;\; : n\geq 0,k\geq 1 \right \} \cup \left \{ a^{n+k}b^{n} \;\; : n\geq 0,k\geq 3 \right \} \end{align*}
Assuming L is regular. Let $m$ be the pumping length given by the pumping lemma.
We pick our string , $w =$ $a^n b^{2n}$ , when $n=k$ and $n,k \geq 3$
then $w \in L$ must be true,
for $|w|>m$ assume n = m,
$w =$ $a^m b^{2m}$ ,
Then from the pumping lemma there exists $x, y, z \in Σ^{∗}$ such that w = xyz , |xy| ≤ m and |y| ≥ 1.
$w_{i} = xy^{i}z$
y must contain entirely of a's
suppose $|y|=k$ and $1 \leq k \leq m$,
then $|xy^{1}z|$ = $a^{m}a^{k}b^{2m}$
if $k = m$ and $i = 1,$
= $|xy^{1}z|$ = $a^{2m}b^{2m}$
This shows that $n_{a} = n_{b}$ but no such string exists in L because in L either $n_{a} > n_{b}$ or $n_{a} < n_{b}$.
This indicates that the assumption that L is regular must be false.
Hence L is not regular