 Multiplicative block-count language

This formal language is not a context-free language. For pumping length $p$, use $a^p{b}^{p^2}{c}^p$. A pumpable window of length at most $p$ cannot meet both outer blocks. If pumping adds $\alpha ,\beta ,\gamma$ copies of $a,b,c$, then $\alpha \gamma =0$ and $1\le \alpha +\beta +\gamma \le p$. Preserving the product constraint would require $\beta =p(\alpha +\gamma )$, impossible both when $\alpha +\gamma =0$ and when it is positive. This contradicts the pumping lemma for context-free languages.