TITLE: Proof of the vector calculus identity $\nabla\cdot(u\times v)=v\cdot(\nabla\times u)-u\cdot(\nabla\times v)$
QUESTION [5 upvotes]: $$\nabla\cdot(u\times v)=v\cdot(\nabla\times u)-u\cdot(\nabla\times v)$$
How can we go about proving the above vector calculus identity component wise?

REPLY [2 votes]: Alternatively, you could use suffix notation:
$$\begin{align*} \boldsymbol{\nabla}\cdot(\mathbf{u}\times\mathbf{v})
&=\partial_i\epsilon_{ijk}u_jv_k \quad\quad\text{definition of Levi-Civita symbol} \\
&=\epsilon_{ijk}\partial_iu_jv_k \quad\quad\text{Levi-Civita symbol is constant} \\
&=\epsilon_{ijk}(v_k\partial_iu_j+u_j\partial_iv_k) \quad\quad\text{derivative of a product} \\
&=\epsilon_{kij}v_k\partial_iu_j-\epsilon_{jik}u_j\partial_iv_k \quad\quad\text{distributivity and property of Levi-Civita symbol} \\
&=v_k\epsilon_{kij}\partial_iu_j-u_j\epsilon_{jik}\partial_iv_k \quad\quad\text{commutativity} \\
&=\mathbf{v}\cdot(\boldsymbol{\nabla}\times\mathbf{u})-
\mathbf{u}\cdot(\boldsymbol{\nabla}\times\mathbf{v}) \quad\quad\text{definition of Levi-Civita symbol}\end{align*}$$