 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/1/i/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} .

The cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two curls proves

\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B) .

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