= Solution
\b[The <formal language> is not a <context-free language>.] Suppose otherwise, and let $p\geq1$ be a <pumping length> from the <pumping lemma for context-free languages>. Choose
$$
w=a^p b^{p^2}c^p.
$$
This <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 give
$$
w=uvxyz,\qquad |vxy|\leq p,\qquad |vy|>0,
$$
with $uv^ixy^iz$ in the <formal language> for every <integer> $i\geq0$.
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 $p$. Hence
$$
\alpha\gamma=0,\qquad 1\leq\alpha+\beta+\gamma\leq p.
$$
Pump 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, is
$$
p^2+\beta=(p+\alpha)(p+\gamma),
\quad\text{or equivalently}\quad
\beta=p(\alpha+\gamma),
$$
where $\alpha\gamma=0$ was used.
If $\alpha+\gamma=0$, then $\beta>0$, contradicting that equality. If $\alpha+\gamma\geq1$, then
$$
\beta\leq p-(\alpha+\gamma)<p\leq p(\alpha+\gamma),
$$
again 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.
$$
\boxed{\{a^m b^{mn}c^n:m,n\geq0\}\text{ is not context-free}.}
$$
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