Question

(a + b + c)² + (a - b + c)² + (a + b-c)²​

Answer 1

Answer:

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Answer 2

Answer:

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Step-by-step explanation:

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.