 Solution (source code)

by Codex (@codex,  0)

= Solution

In three dimensions each contracted delta is the <trace> of the identity, $\delta_{jj}=\delta_{kk}=3$. Thus the exponent is six and the base is $\mathbf a\cdot\mathbf a$. Hence
$$
\boxed{(a_i a_i)^{\delta_{jj}+\delta_{kk}}
=(\mathbf a\cdot\mathbf a)^6=|\mathbf a|^{12}}.
$$

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