Proving jacobi identity with Kronecker Delta and Levi Civita

Question

I am stuck with this task. Maybe someone of you can help. I dont even know where to start.
Unfortunately it is in german, but my heading is basically the question.

Task 1:

Task 10 + Solution:

Answer 1

The $i$-th component of $a \times (b \times c)$ is \begin{align} [a \times (b \times c)]_i &= \sum_{j,k} \epsilon_{ijk} a_j [b \times c]_k \\ &=\sum_{j,k} \epsilon_{ijk} a_j \sum_{m,n} \epsilon_{kmn} b_m c_n \\ &= \sum_{j,m,n} \left(\sum_k \epsilon_{kij} \epsilon_{kmn}\right)a_j b_m c_n \\ &= \sum_{j,m,n} \delta_{im} \delta_{jn} a_j b_m c_n - \sum_{j,m,n} \delta_{in} \delta_{jm} a_j b_m c_n \\ &= b_i \sum_j a_j c_j - c_i \sum_j a_j b_j \\ &= b_i (a\cdot c) - c_i (a \cdot b), \end{align} where we used $\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij}$ and the identity $$\sum_k \epsilon_{kij} \epsilon_{kmn} = \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm}.$$ Therefore $$a \times (b \times c) = b (a\cdot c) - c (a\cdot b),$$ and if we apply this three times we simply get that

\begin{align} &a \times (b \times c) + b \times (c \times a) + c \times (a \times b) \\ & = b (a\cdot c) - c (a\cdot b) + c (b\cdot a) - a (b\cdot c) + a (c\cdot b) - b (c\cdot a) = 0. \end{align}