Tensor notation of a triple scalar product 
I want to write the tensor notation for $$[a\dot\ (b\times c)]a=(a\times b)\times (a\times c).$$

What I got so far is:
$$a \dot\ (b\times c)=a_i(\epsilon_{ijk}b_jc_k)=\epsilon_{ijk}a_ib_jc_k.$$ Now If I let, $F=a\times b=F_i\hat{e}_i$ and $E=a\times c=E_i\hat{e}_i.$ Then, $$F_i=\epsilon_{ijk}a_jb_k$$ $$E_i=\epsilon_{ilm}a_lc_m$$ $$\left[\epsilon_{ijk}a_ib_jc_k\right]a=\left(\epsilon_{ijk}a_jb_k\right)\times \left(\epsilon_{ilm}a_lc_m\right).$$
But I am stuck now because I do not know how to write the tensor notation of a cross product of a cross product. I know I must use the identity $$\epsilon_{ijk}\epsilon_{k\ell m} = \delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}$$ but I do not know how to apply it. 
 A: You can use the following definition of the cross product
$$
a \times b = \epsilon_{ijk} a_j b_k \hat{e}_i
$$
So your second cross product $(a \times b) \times (a \times c) = $ is
$$
\epsilon_{ijk} a_j b_k \hat{e}_i \times \epsilon_{lmn} a_m c_n \hat{e}_l \\
= \epsilon_{rst} (\epsilon_{ijk} a_j b_k \hat{e}_i \cdot \hat{e}_s)(\epsilon_{lmn} a_m c_n \hat{e}_l \cdot \hat{e}_t) \hat{e}_r
$$
The second step merely expresses the fact that the cross-product formula requires the s-th component of the first vector and the t-th component of the second.
Now use $\hat{e}_i \cdot \hat{e}_s = \delta_{is}$ etc
$$
(a \times b) \times (a \times c) = \\
\epsilon_{rst} (\epsilon_{ijk} a_j b_k \delta_{is})(\epsilon_{lmn} a_m c_n \delta_{lt}) \hat{e}_r \\
= \epsilon_{rst} \epsilon_{sjk} \epsilon_{tmn} a_j b_k a_m c_n \hat{e}_r 
$$
Now use the $\epsilon-\delta$ identity $\epsilon_{trs}\epsilon_{tmn} = \delta_{rm}\delta_{sn} - \delta_{rn}\delta_{sm}$ and simplify this further. You have 
$$
(a \times b) \times (a \times c) = \\
\epsilon_{sjk}\delta_{rm}\delta_{sn}a_j b_k a_m c_n \hat{e}_r - \epsilon_{sjk}\delta_{rn}\delta_{sm}a_j b_k a_m c_n \hat{e}_r \\
=\epsilon_{sjk}a_j b_k a_r c_s \hat{e}_r - \epsilon_{sjk}a_j b_k a_s c_r \hat{e}_r \\
$$
Since $\epsilon_{sjk} c_s a_j b_k = c \cdot (a \times b)$, and $\epsilon_{sjk} a_j b_k a_s = 0$, your final expression for $(a \times b) \times (a \times c) = $ is 
$(c \cdot (a \times b)) a_r \hat{e}_r$ = $(c \cdot (a \times b)) a$. 
