 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/1/a/solution

\nabla \cdot (a \times b) = \partial_{i} (ϵ_{ijk} a_{j} b_{k}) = ϵ_{ijk} (\partial_{i} a_{j}) b_{k} + ϵ_{ijk} a_{j} \partial_{i} b_{k} = b_{k} ϵ_{kij} \partial_{i} a_{j} - a_{j} ϵ_{jik} \partial_{i} b_{k} = b \cdot (\nabla \times a) - a \cdot (\nabla \times b) .

The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

a \cdot (\nabla \times b) = b \cdot (\nabla \times a) - \nabla \cdot (a \times b) .

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