Identity proof using levi civita

Question

I struggle to proof \begin{align*} (\vec{a}\times \vec{b})\cdot [(\vec{b}\times \vec{c})\times(\vec{c}\times \vec{a})] = [\vec{a}\cdot (\vec{b} \times \vec{c})]^2 \end{align*}

using Levi-Civita as I'm new to this topic.

I started with \begin{align*} (\vec{a}\times \vec{b})\cdot [(\vec{b}\times \vec{c})\times(\vec{c}\times \vec{a})] &= \varepsilon_{ijk}a_ib_j \cdot [(\varepsilon_{lmn}b_nc_m) \times (\varepsilon_{pqr}c_pa_q)] \end{align*} but I'm unsure about how to continue. Every attempt ended in no man's land. Can someone please help me or give me hints?

Answer 1

\begin{eqnarray} (a\times b)\cdot[(b \times c)\times (c\times a)] &=& (\epsilon_{ijk}a_i b_j \hat{e}_k)\cdot[(\epsilon_{lmn}b_l c_m \hat{e}_n)\times (\epsilon_{pqr}c_p a_q\hat{e}_r)] \\ &=&(\epsilon_{ijk}a_i b_j \hat{e}_k)\cdot [\epsilon_{n r s}(\epsilon_{lmn}b_l c_m )(\epsilon_{pqr}c_p a_q)\hat{e}_s] \\ &=& (\epsilon_{ijs}a_i b_j) \epsilon_{n r s}(\epsilon_{lmn}b_l c_m )(\epsilon_{pqr}c_p a_q) \\ &=& (\epsilon_{ij\color{red}{s}}\epsilon_{nr\color{red}{s}})(\epsilon_{lmn}\epsilon_{pqr})(a_i b_j b_l c_m c_p a_q) \\ &=& (\delta_{in}\delta_{jr}-\delta_{ir}\delta_{jn})(\epsilon_{lmn}\epsilon_{pqr})(a_i b_j b_l c_m c_p a_q) \\ &=& \epsilon_{lmi}\epsilon_{pqj} (a_i b_j b_l c_m c_p a_q) - \epsilon_{lmj}\epsilon_{pqi} (a_i b_j b_l c_m c_p a_q) \\ &=& (\epsilon_{lmi} a_ic_m b_l)(\epsilon_{pqj}a_q b_j c_p) - \underbrace{(\epsilon_{\color{red}{l}m\color{red}{j}}b_\color{red}{l}b_\color{red}{j})}_{=0}\underbrace{(\epsilon_{p\color{orange}{q}\color{orange}{i}}a_\color{orange}{i} a_\color{orange}{q})}_{=0}(c_m c_p) \\ &=& (\epsilon_{ilm} a_ib_lc_m)(\epsilon_{qjp}a_q b_j c_p) \\ &=& [a\cdot (b\times c)][a\cdot (b\times c)] \\ &=& \color{blue}{[a\cdot (b\times c)]^2} \end{eqnarray}

Answer 2

Let $d=b\times c$. The left-hand side is $$\epsilon_{ijk}\epsilon_{ilm}\epsilon_{mpq}a_j b_kd_lc_p a_q=(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})\epsilon_{mpq} a_j b_kd_lc_p a_q.$$The second term vanishes due to an $\epsilon_{jpq} a_j a_q$ factor, so the result is $$\epsilon_{kpq} a_j b_k d_j c_p a_q=[b\cdot (c\times a)][a\cdot d]=[a\cdot(b\times c)]^2.$$