Question

A^2+b^2+c^2-ab-bc-ca =0 then prove that a=b=c

Answer 1

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 Equ. (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).

Answer 2

Given,

a²+b²+c²-ab-bc-ca = 0

Multiply 2 on both sides,

2(a²+b²+c²-ab-bc-ca) = 2(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

(a-b)² = (b-c)² = (c-a)² = 0

(a-b)² = 0

a = b

(b-c)² = 0

b = c

(c-a)² = 0

c = a

Therefore, a=b=c

Hence proved.

Hope it helps