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📅 May 3, 2026 👤 scraface

May 03, 2026

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📌 x \in (-\infty, 2) \cup \left(rac{11}{3}, \infty

📌 oxed{(-\infty, 2) \cup \left(rac{11}{3}, \infty

📌 Question: Let $ a, b, c $ be real numbers such that $ a + b + c = 0 $. Prove that $ a^3 + b^3 + c^3 = 3abc $.

📌 Solution: We use the well-known algebraic identity:

📌 a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).

📌 Given that $ a + b + c = 0 $, substitute into the identity:

📌 a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - bc - ca) = 0.

📌 a^3 + b^3 + c^3 = 3abc.

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