Question

Prove (A×B) ×( C ×D) = [A ∙ (C ×D)]B − [B ∙( C ×D ] A

Answer 1

(A×B) ×( C ×D) = {in next line, reverse cross prod so reverse sign}
- ( C ×D) x (A×B) = {temp sub V = CxD}
- V x ( A x B ) = {next line is from triple vector prod}
- { [V.B] A - [V.A] B } =
- [(C ×D).B] A + [(C ×D).A] B = {next line, just re-order}
[(C ×D).A] B - [(C ×D).B] A = {next line follows from def of triple scalar prod}
[A ∙ (C ×D)]B − [B ∙( C ×D ] A

Answer 2

If we use the tensor calculus, namely we use the Ricci's tensor:
\[\Large {\varepsilon _{ijk}}\]
then we can write the subsequent steps:
\[\Large \begin{gathered}
{\left[ {\left( {A \times B} \right) \times \left( {C \times D} \right)} \right]_i} = {\varepsilon _{ijk}}{\left( {A \times B} \right)_j}{\left( {C \times D} \right)_k} = \hfill \\
\hfill \\
= {\varepsilon _{ijk}}{\varepsilon _{jlm}}{A_l}{B_m}{\varepsilon _{kpq}}{C_p}{D_q} = \left( { - {\varepsilon _{jik}}{\varepsilon _{jlm}}{A_l}{B_m}} \right){\varepsilon _{kpq}}{C_p}{D_q} = \hfill \\
\hfill \\
= \left\{ { - \left( {{\delta _{il}}{\delta _{km}} - {\delta _{im}}{\delta _{kl}}} \right){A_l}{B_m}} \right\}{\varepsilon _{kpq}}{C_p}{D_q} = \hfill \\
\hfill \\
= \left( { - {A_i}{B_k} + {A_k}{B_i}} \right){\varepsilon _{kpq}}{C_p}{D_q} = \hfill \\
\hfill \\
= - {B_k}\left( {{\varepsilon _{kpq}}{C_p}{D_q}} \right){A_i} + {A_k}\left( {{\varepsilon _{kpq}}{C_p}{D_q}} \right){B_i} = \hfill \\
\hfill \\
= - {B_k}{\left( {C \times D} \right)_k}{A_i} + {A_k}{\left( {C \times D} \right)_k}{B_i} = \hfill \\
\hfill \\
= \left\{ { - B \cdot \left( {C \times D} \right)} \right\}{A_i} + \left\{ {A \cdot \left( {C \times D} \right)} \right\}{B_i} \hfill \\
\end{gathered} \]
where the sum is over repeated indices, and:
\[\Large {\delta _{ij}}\]
is the Kronecker symbol
Furthermore, I have used this identity:
\[\Large {\varepsilon _{jik}}{\varepsilon _{jlm}} = {\delta _{il}}{\delta _{km}} - {\delta _{im}}{\delta _{kl}}\]
Finally, the i-th component of vector B is indicated as below:
\[\Large {B_i}\]