Product of two Levi-Civita permutation symbols I am attempting to follow the solution provided by (Mark Viola) here (Kronecker delta and Levi-Civita symbol). I do not understand the following: Step 1 & 2
where the initial Levi-Civita symbols are expanded from only 1 term into 3.
I am assuming this is a lack of dot product and/or cross product knowledge. Any help would be appreciatetd.
 A: We begin with the identity
$$\begin{align}
\varepsilon_{abc}\varepsilon_{k\ell m}&=(\hat x_a\cdot(\hat x_b\times \hat x_c))(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\
\tag1
\end{align}$$
In $(1)$, all indices are fixed.
Now, there are two cases.  In Case $(i)$, none of the indices $(k,\ell,m)$ is equal to either of the other two and hence the number $a$ is equal to one of $k$, $\ell$, or $m$.  In Case $(ii)$, two (or all three) of the indices $(k,\ell,m)$ are equal, in which case $\varepsilon_{k\ell m}=0$.
In Case $(i)$, we can write $\hat x_a$ as the sum
$$\hat x_a=\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\tag 2$$
where $\delta_{ij}$ is the Kronecker Delta and we are not summing over the indices.
Note that we can substitute $(2)$ in $(1)$ in both Cases $(i)$ and $(ii)$ since for Case $(ii)$, $\varepsilon_{k\ell m}=0$.  Therefore, enforcing this substitution reveals
$$\begin{align}
\varepsilon_{abc}\varepsilon_{k\ell m}&=\left(\left(\delta_{ak}\hat x_k+\delta_{a\ell}\hat x_\ell+\delta_{am}\hat x_m\right)\cdot(\hat x_b\times \hat x_c)\right)(\hat x_k\cdot(\hat x_\ell\times \hat x_m))\\\\
\tag3
\end{align}$$
Finally, exploiting the property of the scalar triple product
$$\hat x_k\cdot (\hat x_\ell\times \hat x_m)=\hat x_\ell\cdot (\hat x_m\times \hat x_k)=\hat x_m\cdot (\hat x_k\times \hat x_\ell)$$
leads to the coveted eqality
$$\begin{align}
\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)&=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot \hat x_k\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\\\\
&+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)\\\\
&+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_k\times \hat x_{\ell})\right)
\end{align}$$
as was to be shown!
