Let a, b, c be real numbers, not all of them are equal. Prove that if

1.	Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.
Answer» I. a² + ab + b² = (– b – c)² + (– b – c) b + b² {as a + b + c = 0, a = – b – c} = (b² + c² + 2bc) – b² – bc + b² = b² + c² + bc Similarly b² + c² + bc = c² + ca + a² ∴ a² + ab + b² = b² + c² + bc = c² + a² + ac. II. a² + ab + b² = b² + bc + c² a²+ ab – bc – c² = 0 a² – c² + b (a – c) = 0 (a – c) (a + c + b) = 0 a – c = 0 or a + b + c = 0 as a – c = 0 is not possible as a, b, c are not equal. ∴ a + b + c = 0.

Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.

Answer»

I. a² + ab + b² = (– b – c)² + (– b – c) b + b² {as a + b + c = 0, a = – b – c}

= (b² + c² + 2bc) – b² – bc + b²

= b² + c² + bc

Similarly b² + c² + bc = c² + ca + a²

∴ a² + ab + b² = b² + c² + bc = c² + a² + ac.

II. a² + ab + b² = b² + bc + c²

a²+ ab – bc – c² = 0

a² – c² + b (a – c) = 0

(a – c) (a + c + b) = 0

a – c = 0 or a + b + c = 0

as a – c = 0 is not possible as a, b, c are not equal.

∴ a + b + c = 0.

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Let a, b, c be real numbers, not all of them are equal. Prove that if a + b + c = 0, then a2 + ab + b2 = b2 + bc + c2 = c2 + ca + a2.Prove the converse, if a2 + ab + b2 = b2 + bc + c2 = c2 = ca + a2, then a + b + c = 0.

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