Question

if a²+b²+c²-ab-bc-ca=0,then show that a=b=c​

Answer 1

$\huge\underline\bold\red{AnswEr}$

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.

Answer 2

Q. if a²+b²+c²-ab-bc-ca=0,then show that a=b=c

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$\huge\bold\red{AnsweR:-}$

a² + b² + c² = ab + bc + ca

By 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

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

That proves : a = b = c.

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