= Solution
Use <suffix notation>, summing repeated indices, and the <Levi-Civita symbol> convention $(A\times B)_i=\epsilon_{ijk}A_jB_k$. Contracting the outer symbol with the symbol from the second <cross product> gives the <epsilon-delta identity>
$$
\epsilon_{ijk}\epsilon_{kpq}=\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp},
$$
where $\delta$ is the <Kronecker delta>. Thus
$$
\begin{aligned}
\bigl((A\times B)\times(A\times C)\bigr)_i
&=\epsilon_{ijk}\epsilon_{j\ell m}\epsilon_{kpq}A_\ell B_mA_pC_q\\
&=(\delta_{ip}\delta_{jq}-\delta_{iq}\delta_{jp})\epsilon_{j\ell m}A_\ell B_mA_pC_q\\
&=A_i\epsilon_{q\ell m}C_qA_\ell B_m
-C_i\epsilon_{p\ell m}A_pA_\ell B_m\\
&=A_i\epsilon_{\ell mq}A_\ell B_mC_q.
\end{aligned}
$$
The second term is zero because $A_pA_\ell$ is symmetric in $p,\ell$ and the <Levi-Civita symbol> is antisymmetric. The cyclic rearrangement of the remaining symbol has positive sign. Its contraction is the <scalar triple product> $A\cdot(B\times C)$. Hence
$$
\boxed{(A\times B)\times(A\times C)=(A\cdot(B\times C))A.}
$$
This <cross products with a common vector> identity requires no independence assumption, so it also covers zero or parallel vectors.
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