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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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.
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.
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.