Question

if a^2+b^2+c^2-ab-bc-ca=0,prove that a=b=c​

Answer 1

Answer:

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

Answer 2

a² + b² + c² - ab - bc - ca = 0

multiplying throughout by 2

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 = 0 ... (if x² + y² + z² = 0, x = y = z = 0)

b - c = 0

c - a = 0

a = b ... (i)

b = c ... (ii)

c = a ... (iii)

from (i), (ii) and (iii)

a = b = c

... Hence Proved!