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If a^2+b^2+c^2-ab--bc-ca=0 then find the value of a^3+b^3+c^3= |
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Answer» a² + b² + c² = ab + BC + ca On MULTIPLYING both sides by “2”, it becomes 2 ( a² + b² + c² ) = 2 ( ab + bc + ca) 2a² + 2b² + 2c² = 2AB + 2bc + 2ca a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0 a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0 (a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0 (a – b)² + (b – C)² + (c – a)² = 0 => Since the sum of SQUARE is zero then each term should be zero ⇒ (a –b)² = 0, (b – c)² = 0, (c – a)² = 0 ⇒ (a –b) = 0, (b – c) = 0, (c – a) = 0 ⇒ a = b, b = c, c = a ∴ a = b = c. |
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