Proving the given language isn't context free using Pumping Lemma i want to prove that the following Language isn't CF

$L=\{a^kba^kba^k|k\in \mathbb{N}\}$

Let $z=a^nba^nba^n$ be a String from $L$

$n$ is the pumping length and
$|z|=3n+2 &gt; n$

and $z=uvwxy$ with $|vx|\geqslant 1$, $|vwx|\leqslant n$

but i'm not sure how to determine the correct substrings, how should i consider  them ? I tried the ones below are thy correct ?

Case 1 :
$u=a^n,\ uwx=ba^n,\ y=ba^n$

Case 2 :
$u=a^{n-l},\ uwx=a^lb,\ y=a^nba^n$

Case 3 :
$u=a^nb,\ uwx=a^nb,\ y=a^n$

Case 4 :
$u=a^nba^{n-l},\ uwx=a^{l-k},\ y=a^kba^n$

Case 5 :
$u=a^nb,\ uwx=a^n,\ y=ba^n$
The pumping segments $v$ and $x$ can be anywhere in the string assuming they are at most $n$ apart, i.e., $|vwx|\le n$. Like you I get five possibilities, but I think you wrote some special cases rather than the general picture.

Inside one of the $a$-segments, for instance :


the first $u=a^i$, $vwx=a^j$, $y=a^{n-i-j}ba^nba^n$ where $i+j\le n$ and $j\ge 1$. Or  
the second $u=a^nba^i$, $vwx=a^j$, $y=a^{n-i-j}ba^n$ where $i+j\le n$ and $j\ge 1$. Or 
the third $\dots$ .


Or overlapping with one of the $b$'s. Of course because of the lengths we cannot overlap with both $b$'s.


First: $u=a^{n-i}$, $vwx=a^iba^j$, $y=a^{n-j}ba^n$ where $0\le i,j\le n$, $i+j&lt; n$ .
Second $\dots$ .


However this does not solve the pumping problem. Note we repeat the strings $v,x$ when pumping. In these last cases we do not know how $v,x$ look. Do they contain the $b$? More cases.

A shortcut to the proof would be to make the following observation.


The pumping segments cannot contain $b$.


That leaves five cases for the position of $v$, $x$ in the three $a$-segments. Either deal with them or argue


Now that $v$, $x$ can only contain $a$'s we can pump either one or two of the three $a$-segments in $z$. That means after pumping they cannot all contain the same number of $a$'s