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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