If a2+b2+c2=ab+bc+ca find a+b/b

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.

hence

a+b/b

a+a/a= (since a=b)

2a/a=

a

54

step-by-step explanation:

6x9

your answer is a

step-by-step explanation:

4*4*4*4=256