Prove using pumping free lemma for context-free languages
i figured it out

$w = (ac^pb)^p $ cannot be split.

in the case that vxy consists of only a's and b's $uv^0xy^0z $ would have less a's and b's but not less c's.

in the case that vxy consists of only c's $uv^0xy^0z $ would have less c's but not less a's and b's.

in the case of a combination it might have an a or a b or both. if it has an a, the number of c's needs to increase by p, but $|vy| \le p$ and we'd need p + 1 terminals. for b the same holds, and for an a and a b it would need even more c's.

so we cannot split this word into 5 substrings  u,v,x,y,z such that $uv^ixy^iz$ is in L