Working in index notation, the $i$th component of the left side is: \begin{align*} [a \times (b \times c)]_i &= \varepsilon_{ijk} a_j (b \times c)_k = \varepsilon_{ijk} a_j \varepsilon_{kpq} b_p c_q. \end{align*} Rearranging indices and applying the [Epsilon-Delta Identity](/theorems/913) with the contraction on $k$: \begin{align*} \varepsilon_{ijk}\varepsilon_{kpq} = \varepsilon_{kij}\varepsilon_{kpq} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}. \end{align*} (The first step uses $\varepsilon_{ijk} = \varepsilon_{kij}$ by cyclic symmetry.) Substituting: \begin{align*} [a \times (b \times c)]_i &= (\delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp}) a_j b_p c_q = a_j b_i c_j - a_j b_j c_i \\ &= (a \cdot c) b_i - (a \cdot b) c_i. \end{align*} This is the $i$th component of $(a \cdot c)b - (a \cdot b)c$.