How can I prove, using the pumping lemma for context free languages, that $\{ll^{R}l|l\in\{a,b\}^{*}\}$is not a context free language ?
I tried to put $n$ as the pumping lemma constant and chose $a^{n}a^{n}a^{n}$ but this does seem to have a factorization since all I need is that the amount of $a$ in the word is divisible by $3$.
I then tried to take $l':=a^{n}b^{n}b^{n}a^{n}a^{n}b^{n}$ ($l=a^{n}b^{n})$, I am having a hard time even figuring out if this will lead to a contradiction:
My thoughts are that if $l'=uvwxy$ then since $|vwx|\leq n$ it must be contained in the $a^{n}$ or $b^{n}$ or be something like $a^{k_{1}}b^{k2}$ or $b^{k_{1}}a^{k2}$. but I'm having a difficult time continue from here, it seems that now I should go over each option and devide to options of how to factor it to $v,w,x$ which gives many cases.
Am I on the right path or can I do something easier/smarter ?
Please note: Since this is important to me, as this is how I prepare for my exam tomorrow (thus can't offer a bounty right now), I will give $50$ points (+upvote+accept) for a somewhat full solution of this question