Proving vector Identities (Using the Permutation Tensor and Kroenecker Delta) Prove the following vector identities by using permutation tensor and kroenecker delta.
$$(\vec{A} \times \vec{B}) \times (\vec{C} \times \vec{D}) = (\vec{A} \cdot (\vec{B} \times \vec{D})) \cdot \vec{C} - (\vec{A} \cdot (\vec{B} \times \vec{C})) \cdot \vec{D}$$
I started by assuming $\vec{F}=\vec{A} \times \vec{B} = \varepsilon_{ijk} A_j B_j$ and $\vec{E}=\vec{C} \times \vec{D} = \varepsilon_{mnp} A_n B_p$
Then
$\vec{G} = \vec{F} \times \vec{E} = \varepsilon_{qim} F_i E_m = \varepsilon_{qim} \varepsilon_{ijk} \varepsilon_{mnp}A_j B_k C_n D_p$
After this I have dont know what else to do. Could be please give me some tipps.
Thanks.
 A: Since no one has answered yet, I'm assuming your still having trouble with it.
The trick for your question was to realise
$$\begin{align}
\implies [(\vec A \times \vec B) \times (\vec C \times \vec D)]_{i} &= [-(\vec B \times \vec A) \times (\vec C \times \vec D)]_{i} \\
&= [(\vec C \times \vec D) \times (\vec B \times \vec A)]_{i}
\end{align}$$
Set
$$\begin{align}
\vec E &= (\vec C \times \vec D) \\
\vec F &= (\vec B \times \vec A) \\
\end{align}$$
Hence
$$\begin{align}
[\vec E \times \vec F]_{i} &= \epsilon_{ijk} E_{j} F_{k} \\
&= \epsilon_{ijk} (\vec C \times \vec D)_{j} \vec F_{k} \\
&= \epsilon_{ijk} [\epsilon_{jlm} C_{l} D_{m}] F_{k} \\
&= \epsilon_{jki} \epsilon_{jlm} C_{l} D_{m} F_{k} \\
&= (\delta_{kl} \delta_{im} - \delta_{km} \delta_{il}) C_{l} D_{m} F_{k} \\
&= (C_{k} F_{k}) D_{i} - (D_{k} F_{k}) C_{i} \\
&= [(C \cdot F) D - (D \cdot F) C]_{i} \\
&= [C \cdot (B \times A) D - D \cdot (B \times A) C] \\
&= [A \cdot (C \times B) D - A \cdot (D \times B) C] \\
&= [A \cdot (-(B \times C)) D - A \cdot (-(B \times D)) C] \ \ (*) \\
&= [A \cdot (B \times D) C - A \cdot (B \times C) D]
\end{align}$$
where at $(*)$ we used the scalar triple product (you can use Levi-Civita symbols to prove this symmetry but I didn't bother as I'm sure you got the gist of the proof).
