Question

question in the form of (a+b+c)(a²+b²+c²-ab-bc-ac) in variables​​

Answer 1

Multiply by 2 (both RHS and LHS)

2( a² + b² + c² - ab - bc - ca ) = 0

=> 2a² +2 b² + 2c² - 2ab - 2bc - 2ca = 0

=> a² + b² - 2ab + b² + c² - 2bc + a² + c² - 2 ca =0

=> (a-b)² + (b-c)² + (c-a)² = 0

Now, sum of three positive numbers can not be zero there fore all the numbers are zero.

=> a = b = c

Answer 2

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.

Hope this will helps u......be happy......