An algebra summation problem Given 1D arrays $A$, $B$, $C$, $D$ and $E$, all of length 3.
Please verify that：
$$
2 \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \sum_{n=1}^{3} \ \epsilon_{ijk} \ A_{n} \ B_{j} \ C_{k} \ D_{n} \ E_{i} + \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \sum_{n=1}^{3} \epsilon_{ijk} \ A_{n} \ \Big(B_{n} \ C_{j} - B_{j} \ C_{n}\Big) \ D_{k} \ E_{i} \\ = \\
2 \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \ \epsilon_{ijk} \ A_{i} \ B_{j} \ C_{k} \ D_{i} \ E_{i} + \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3} \epsilon_{ijk} \ A_{i} \ \Big(B_{i} \ C_{j} - B_{j} \ C_{i}\Big) \Big(D_{k} \ E_{i} + D_{i} \ E_{k}\Big)
$$
where $\epsilon_{ijk}$ is:
$$
\epsilon_{ijk} = \begin{cases}
1 & (i,j,k) \in \Big\{(1,2,3), (3,1,2), (2,3,1)\Big\}\\
-1 & (i,j,k) \in \Big\{(3,2,1), (2,1,3), (1,3,2)\Big\} \\
0 & \text{otherwise}
\end{cases}
$$
PS.
I have expanded the terms by brute force, and verified the equation in python program (https://github.com/declmal/verify/blob/master/verify.py), but it's tedious.
I'm still wondering if any simple analytical proof exists.
 A: Let $A,B,C$ be three column vectors having each three components. Then i will write $\det(A,B,C)$ for the determinant of the $3\times 3$ matrix built from the three vectors as columns. The six terms determinant formula over permutations of the three indices, involving their signs which correspond to the $\epsilon$-signs from the OP (Sarrus) reads then
$$
\det(A,B,C)
=\sum_{1\le i,j,k\le 3}\epsilon_{ijk}\; A_iB_jC_k\ .
$$
The given identity can be rewritten to involve determinants (and simple linear forms). For an easy writing, i will use the componentwise product $AB$ for two given vectors $A,B$, so the components of $AB$ are $(AB)_i=A_iB_i$.
When an expression like $ABC$ is used, then it is $ABC:=(AB)C$.
For a vector $A$ i will denote by $s(A)$ the sum of its components.
Then we have to show red equality:
$$
2s(AD)\det(E,B,C)
\\
\qquad+ s(AB)\det(E,C,D)
- s(AC)\det(E,B,D)
\\[4mm]
{\color{red}{\overset ?
=}}\ 
2\det(ADE,B,C)
\\
\qquad+
\det(ABE, C,D)
+
\det(ABD, C,E)
-
\det(ACE, B,D)
-
\det(ACD, B,E)
\ .
$$
We observe that the relation (i.e. the difference expression $\Delta =LHS - RHS$) is linear in each of the vectors $A,B,C,D,E$, (So if the relation is true for $A=A$, $A=A''$ substituted in the position $A$, then it is also true for $A=A'+A''$, this is additivity in $A$, we also have compatibility with multiplication with a scalar in the variable $A$, etc.)
The relation to be shown is antisymmetric in $B,C$. So if $B=C$ the difference $\Delta$ is zero. So it is enough to verify the expression for $B$, $C$ two different canonical basis vectors of $\Bbb R^3$. We use  the column vectors $B=[0,1,0]^T$ and
$C=[0,0,1]^T$. By symmetry (permutation of coordinates in $\Bbb R^3$) this is enough. Then the relation to be shown is:
$$
2s(AD) E_1
\\
\qquad 
- A_2\det\begin{bmatrix} E_1&D_1\\ E_2 &D_2\end{bmatrix}
- A_3\det\begin{bmatrix} E_1&D_1\\ E_3 &D_3\end{bmatrix}
\\[4mm]
{\color{red}{\overset ?
=}}\ 
2A_1D_1E_1
\\
\qquad
+ A_2E_2D_1
+ A_2D_2E_1
-A_3E_3D_1
- A_3D_3E_1
\ .
$$
The relation is still linear in the remained vectors $A,D,E$, so it is enough to verify it for the three chances of $A$ in the canonical basis. We verify it in the cases:

*

*$A=[A_1,0,0]^T=[1,0,0]^T$. Yes, $2A_1D_1E_1$ on both sides.

*$A=[0,A_2,0]^T=[0,1,0]^T$. Yes, the parts in $A_2$ on both sides correspond. The terms in $A_2$ in the LHS are plus $2A_2D_2E_1$, and minus $A_2(E_1D_2-D_2E_1)$. In total $A_2(E_1D_2+D_2E_1)$

*$A=[0,0,A_3]^T=[0,0,1]^T$. Here i see a difference.


(I'm sorry, have to submit with the above conclusion. There were some edits of the question, it would be thus possible that the indices are not fixed. If time allows i will look again on the equality and double check the typed computations. A readable solution (prove/disprove) would go this way. The python code is unfortunately not easy to digest for me. I may try sage for a check if the issue does matter for some specific purpose.)
