Use the 
notation and derive the following two identities:
where 
denotes the triple scalar product 
.
Solution.
In attempting the following proof, extensive reference was made to the article found on the Internet (http://www.ucl.ac.uk/~ucappgu/seminars/levi-civita(DOT)pdf):
This completes the proof of identity (1).
For identity (2), I can make use of identity (1) by substituting 
for 
, 
for 
, and 
for , and get the following:
This completes the proof of identity (2).
Improvement on presentation
The 
-component of 
is:
![\begin{aligned} & \quad [\mathbf{A} \times (\mathbf{B} \times \mathbf{C})]_i \\ & = \sum_{jk}\epsilon_{ijk}A_j(\mathbf{B}\times\mathbf{C})_k \\ & = \sum_{jk}\epsilon_{ijk}A_j\sum_{lm}\epsilon_{klm}B_lC_m \\ & = \sum_{jk}\sum_{lm}\epsilon_{ijk}\epsilon_{klm}A_jB_lC_m \\ \dots\, &\textrm{ by } \sum_k \epsilon_{ijk}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\,\dots\\ & = \sum_{j}\sum_{lm}(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})A_jB_iC_m \\ & = \sum_j (A_jB_iC_j - A_jB_jC_i ) \\ & = (\mathbf{A}\cdot \mathbf{C})B_i-(\mathbf{A}\cdot\mathbf{B})C_i \end{aligned}](https://physicspupil.com/wp-content/ql-cache/quicklatex.com-4309c1a80223a35df19eade04c9ac355_l3.svg "Rendered by QuickLaTeX.com")
In compact form resulted identity (1):
