in this question, If a2+b2+c2=ab+bc+ca , then prove that a=b=c?

Question

(a - b)² + (b - c)² + (c - a)² = 0

=>

(a - b)² = (b - c)² = (c - a)² = 0
HOW DOES THIS HAPPEN?"

Answer 1

Note that

(a-b)^2 + (b-c)^2 + (c-a)^2

= a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ac + a^2

= 2(a^2 + b^2 + c^2) - 2(ab + bc + ac).

Since the two expressions in parentheses are equal,

the difference will be zero.

However, in the expression

(a-b)^2 + (b-c)^2 + (c-a)^2,

none of the three squares can be a negative number,

so if their sum is zero, each of them must separately be zero.

That's how they get

(a-b)^2 = 0, (b-c)^2 = 0, (c-a)^2 = 0,

which proves that a-b = 0, b-c = 0, c-a = 0,

and hence a=b=c.

Answer 2

Do you mean this?



a² + b² + c² = ab + bc + ca

2a² + 2b² + 2c² = 2ab + 2bc + 2ca

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

(a - b)² + (b - c)² + (c - a)² = 0



That is the sum of three squares equal to zero. All squares are non-negative, the sum of three squares can be zero only if each square is zero.



(a - b)² = (b - c)² = (c - a)² = 0



(a - b)² = 0

a - b = 0

a = b



Likewise, b = c, and c = a.



a = b = c

Answer 3

Given that, a2+b2+c2 = ab+bc+ca

So,

a = b = c.