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I'm praticing tensor notation, and I want to prove this way that given vectors $A,B,C,D$ then $(A \times B) \times (C \times D) = \det(A,C,D)B - \det(B,C,D)A$, where $\det$ means the triple product. I'm getting something different here, can someone check? Taking coordinates:

$ ((A \times B) \times (C \times D))_i = \epsilon_{ijk} (A \times B)_j (C \times D)_k = \epsilon_{ijk} \epsilon_{jlm} A_l B_m \epsilon_{kno} C_n D_o $

$= \epsilon_{ijk} \epsilon_{kno} \epsilon_{jlm} A_l B_m C_n D_o$

Here I used this nice little identity:

$ = (\delta_{in} \delta_{jo} - \delta_{io} \delta_{jn}) \epsilon_{jlm} A_l B_m C_n D_o $

$= [\delta_{in} \delta_{jo} \epsilon_{jlm} A_l B_m C_n D_o] - [\delta_{io} \delta_{jn} \epsilon_{jlm} A_l B_m C_n D_o]$

Now I thought the following: since $i$ is a free index $\delta_{in}$ exchanges $n$ for $i$ in the first part, and $\delta_{io}$ exchanges $o$ for $i$ in the second part. Then:

$= [\delta_{jo} \epsilon_{jlm} A_l B_m C_i D_o] - [\delta_{jn} \epsilon_{jlm} A_l B_m C_n D_i]$

Now, $\delta_{jo}$ and $\delta_{jn}$ make the double sum collapse into a single one, saving only the terms on which $j = o$ in the first part and $j = n$ on the second. I'll keep the index $j$. Then:

$= \epsilon_{jlm} A_l B_m C_i D_j - \epsilon_{jlm} A_l B_m C_j D_i$

Now we see the definition of the cross products

$= (A \times B)_j D_j C_i - (A \times B)_j C_j D_i$

$= \det(A,B,D) C_i - \det(A,B,C) D_i$

Where is my error?

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There is no mistake, your solution is correct. Look for a variable substitution:

$$A = C' \qquad B = D' \qquad C = A' \qquad D = B'$$

Your result show that: $$ (C'\times D')\times (A' \times B') = \det(C',D',B')A' - \det(C',D',A')B' $$

Let's just forget lexicografic fomallity and rewrite this as

$$ (C\times D)\times(A \times B) = \det(B,C,D)A - \det(A,C,D)B $$

which is the same you wanted first, since $$(C\times D)\times(A \times B) = -(A\times B)\times(C\times D) $$

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