My task is to prove the following vector identity using Einstein notation: $$\vec a \times (\vec b \times \vec c) + \vec b \times (\vec c \times \vec a) + \vec c \times (\vec a \times \vec b) = \vec0$$
This is my attempt:
$$\varepsilon^i_{\ jk}a^j\varepsilon^k_{\ lm}b^lc^m + \varepsilon^i_{\ jk}b^j\varepsilon^k_{\ lm}c^la^m + \varepsilon^i_{\ jk}c^j\varepsilon^k_{\ lm}a^lb^m = \varepsilon^{ijk}\varepsilon_{klm}a_jb^lc^m + \varepsilon^{ijk}\varepsilon_{klm}b_jc^la^m + \varepsilon^{ijk}\varepsilon_{klm}c_ja^lb^m = \varepsilon^{kij}\varepsilon_{klm}a_jb^lc^m + \varepsilon^{kij}\varepsilon_{klm}b_jc^la^m + \varepsilon^{kij}\varepsilon_{klm}c_ja^lb^m = [\delta_i^l\delta_j^m - \delta_i^m\delta_j^l]a_jb^lc^m + [\delta_i^l\delta_j^m - \delta_i^m\delta_j^l]b_jc^la^m + [\delta_i^l\delta_j^m - \delta_i^m\delta_j^l]c_ja^lb^m = a_jb^ic^j - a_jb^jc^i + b_jc^ia^j - b_jc^ja^i+c_ja^ib^j-c_ja^jb^i $$
I am stuck here and unable to move forward. What can I do now according to the rules of the notation?
Remember the einstein summation convention. In you last expression the first term has an individual b, and a summation over a and c. This is the same in the last term. Only difference is which is upper or which is a lower index. Just use the metric tensor and you see they are equal. Analogously for the rest.
So 1st and 6th will cancel, 2 and 3, 4 and 5.