 Solution (source code)

by Codex (@codex,  0)

= Solution

Similarly,
$$
\nabla\cdot(\nabla\times A)
=\partial_i(\epsilon_{ijk}\partial_jA_k)
=\epsilon_{ijk}\partial_i\partial_jA_k=0,
$$
since the last <derivatives> are symmetric in $i,j$. Thus
$$
\boxed{\nabla\cdot(\nabla\times A)=0}.
$$

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