Question

if a,b,c are real numbers and a2+b2+c2-ab-bc-ca=0 then show that a=b=c

Answer 1

Consider, a2 + b2 + c2 – ab – bc – ca = 0
Multiply both sides with 2, we get
2( a2 + b2 + c2 – ab – bc – ca) = 0
⇒ 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0
⇒ (a2 – 2ab + b2) + (b2 – 2bc + c2) + (c2 – 2ca + a2) = 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.

Answer 2

Heya!!☺

Here is your answer.

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a²+b²+c²-ab-bc-ca=0

=> 1\2 × 2(a²+b²+c²-ab-bc-ca) = 0

=> 1\2 (2a²+2b²+2c²-2ab-2bc-2ca) = 0

=> 1\2 {( a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)}=0

=> 1\2 { ( a-b)² + (b-c)² + (c-a)²} = 0

=> 1\2 × 2 { (a-b)² + (b-c)² + (c-a)²} = 0 × 2

=> (a-b)² + (b-c)² + (c-a)² = 0 -----(1)

From equation 1,

We can say that the value of equation (1) is 0 when,all squares is 0 because
any negative number's square is positive and their sum ≠ 0

So, the only way to have '0' is,

=> (a-a)² + (b-b)² + (c-c)² = 0

So, a = b = c. (Proved)