= Solution
Use <suffix notation>, with repeated indices summed from one to three. The <cross product>, <divergence>, and <curl> have components $(\mathbf A\times\mathbf B)_i=\epsilon_{ijk}A_jB_k$, $\nabla\cdot\mathbf A=\partial_iA_i$, and $(\nabla\times\mathbf A)_i=\epsilon_{ijk}\partial_jA_k$. Here $\epsilon$ is the <Levi-Civita symbol>.
The <product rule> gives
$$
\partial_i(\epsilon_{ijk}A_jB_k)=\epsilon_{ijk}(\partial_iA_j)B_k+\epsilon_{ijk}A_j\partial_iB_k.
$$
Using the cyclic identity $\epsilon_{ijk}=\epsilon_{kij}$ in the first term and $\epsilon_{ijk}=-\epsilon_{jik}$ in the second identifies these terms as $B_k(\nabla\times\mathbf A)_k$ and $-A_j(\nabla\times\mathbf B)_j$. Therefore
$$
\boxed{\nabla\cdot(\mathbf A\times\mathbf B)=\mathbf B\cdot(\nabla\times\mathbf A)-\mathbf A\cdot(\nabla\times\mathbf B).}
$$
For the second identity, contract two <Levi-Civita symbols>, using $\epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$:
$$
\begin{aligned}
[\nabla\times(\mathbf A\times\mathbf B)]_i
&=\epsilon_{ijk}\epsilon_{klm}\partial_j(A_lB_m)\\
&=\partial_j(A_iB_j-A_jB_i)\\
&=B_j\partial_jA_i+A_i\partial_jB_j-A_j\partial_jB_i-B_i\partial_jA_j.
\end{aligned}
$$
The four component expressions translate into
$$
\boxed{\nabla\times(\mathbf A\times\mathbf B)=(\mathbf B\cdot\nabla)\mathbf A-\mathbf B(\nabla\cdot\mathbf A)-(\mathbf A\cdot\nabla)\mathbf B+\mathbf A(\nabla\cdot\mathbf B).}
$$
Both derivations require only continuously differentiable <vector fields>; no additional divergence-free assumptions are used.
Back to article page
