= Solution
Use <Einstein summation convention>, writing $\partial_j=\partial/\partial x_j$. The <contraction of two Levi-Civita symbols> gives
$$
\begin{aligned}
[\nabla\times(\mathbf a\times\mathbf b)]_i
&=\epsilon_{ijk}\partial_j(\epsilon_{klm}a_lb_m)\\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_j(a_lb_m)\\
&=\partial_j(a_ib_j-a_jb_i)\\
&=a_i\partial_jb_j-b_i\partial_ja_j+b_j\partial_ja_i-a_j\partial_jb_i.
\end{aligned}
$$
The last equality is the <product rule>. Recognizing the <divergences> and <directional derivatives> proves the <curl of a cross product> identity:
$$
\boxed{\nabla\times(\mathbf a\times\mathbf b)=\mathbf a(\nabla\cdot\mathbf b)-\mathbf b(\nabla\cdot\mathbf a)+(\mathbf b\cdot\nabla)\mathbf a-(\mathbf a\cdot\nabla)\mathbf b.}
$$
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