If a^2+b^2+c^2=ab+bc+ca, find (c+a)/b.

Given: a^2+b^2+c^2-ab-bc-ca=0

Multiply both sides with 2, we get

2(a^2 + b^2 + c^2 – ab – bc – ca) = 0

2a^2 + 2b^2 + 2c^2 – 2ab – 2bc – 2ca = 0

(a^2 – 2ab + b^2) + (b^2 – 2bc + c^2) + (c^2 – 2ca + a^2) = 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.

therefore c+a/b=2

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