 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/12/a/iii/solution

In three dimensions each contracted delta is the trace of the identity,

δ_{jj} = δ_{kk} = 3

. Thus the exponent is six and the base is

a \cdot a

. Hence

(a_{i} a_{i})^{δ_{jj} + δ_{kk}} = (a \cdot a)^{6} = ∣ a ∣^{12} .

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