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# If a²+b²+c² = ab+bc+ca, Find (c+a)/b.

Algebraic identities are equations where the value of the left-hand side of the equation is identically equal to the value of the right-hand side of the equation.

## Answer: If a²+b²+c² = ab+bc+ca, then (c+a)/b = 2.

Let's look into the steps below

**Explanation**:

Given: a²+b²+c² = ab+bc+ca

On multiplying both the sides by ‘2’, we will get

⇒ 2 ( a² + b² + c² ) = 2 ( ab + bc + ca)

⇒ 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

⇒ 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

By rearranging and grouping the terms we get,

⇒ (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ca + a²) = 0

⇒ (a – b)² + (b – c)² + (c – a)² = 0 (Since, (a – b)² = (a² – 2ab + b²))

As the sum of all the three squares is zero thus, each term will be equal to zero

Therefore,

(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

Thus, a = b = c.

Hence,

(c + a)/b = (b+b)/b

= 2b/b

= 2

### So, If a²+b²+c² = ab+bc+ca, (c+a)/b = 2.

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