Question

If a2+b2+c2-ab-bc-ca=0, prove that a=b=c

Answer 1

 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

Answer:

Yes......a = b = c

Step-by-step explanation:

here , a ² + b² + c² – ab – bc – ca = 0

we have to prove a = b= c

now, the solution .....

by , multiplying both sides with 2, we get the result.

2( a ²+ 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

hence, we can say ,(a - b)² = (b - c)² = (c - a)² = 0

therefore,

we can say, (a - b)² = 0 ---------- (1)

(b - c)² = 0 ---------- (2)

(c - a)² = 0 ---------- (3)

therefore, by Simplifying Equation (1), we get

(a - b)² = 0

now Taking the Square Root on both sides, we have

a - b = 0

a = b ---------- (4)

similarly, Simplifying Equ. (2), we get

(b - c)² = 0

Taking Square Root on both sides, we have

b - c = 0

b = c ---------- (5)

again, Simplifying Equ. (3), we have

(c - a)² = 0

Taking Square Root on both sides, we have

c - a = 0

c = a ---------- (6)

now, From Equation No. (4), (5) & (6) , it is proved that

a = b = c

hence a= b= c , proved..........(ANS).

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