Question

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

HOPE U UNDERSTAND

PLS MARK IT AS BRAINLIEST

Answer 2

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).