 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-79/2/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

a, b, c

a + b + c = 0

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

therefore gives

a^{3} + b^{3} + c^{3} = 3 ab c

. Since the trace of

S^{3}

is the sum of the cubes of its eigenvalues, the Betchov relation becomes

3 ⟨ ab c ⟩ = - \frac{3}{4} ⟨ ω_{i} ω_{j} S_{ij} ⟩, ⟨ ab c ⟩ = - \frac{1}{4} ⟨ ω_{i} ω_{j} S_{ij} ⟩ .

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