 Solution

The formal language is not a context-free language. Suppose otherwise, and let $p\ge 1$ be a pumping length from the pumping lemma for context-free languages. ChooseThis word belongs to the formal language, since its middle exponent is the product of the two outer exponents. The pumping lemma for context-free languages would givewith $u{v}^{i}x{y}^{i}z$ in the formal language for every integer $i\ge 0$.

Let $\alpha ,\beta ,\gamma$ be the respective numbers of $a,b,c$ symbols in the two pumped pieces $v,y$ together. The window $vxy$ cannot contain both an $a$ and a $c$: it would have to contain the entire middle block of length $p^2$, plus at least those two outer symbols, exceeding $p$. HencePump once extra, taking $i=2$. The resulting symbol counts are $p+\alpha$, $p^2+\beta$, and $p+\gamma$. A necessary condition for membership, regardless of whether the block order survives, iswhere $\alpha \gamma =0$ was used.

If $\alpha +\gamma =0$, then $\beta >0$, contradicting that equality. If $\alpha +\gamma \ge 1$, thenagain contradicting it. Thus every permissible decomposition fails at $i=2$. This contradicts the pumping lemma for context-free languages and proves the claim. The obstruction is the multiplicative block-count language: a bounded local change cannot preserve the product constraint.