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.
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).