Question

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

Answer 1

Answer:

Now,

a²+b²+c²-ab-bc-ca=0

we have to prove a = b= c

Multiply both sides of the equation with 2. Remember, 2×0=0.

2( a ²+ 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  [because  (a² – 2ab + b²) = (a - b)², (b² – 2bc + c²) = (b - c)², (c² – 2ca + a²) = (c - a)²]

hence, we can say ,(a - b)² = (b - c)² = (c - a)² = 0 , because a² + b² + c² =0 means a²=0, b²=0, c²=0.

therefore,

we can say, (a - b)² = 0..........(1)

(b - c)² = 0.......................(2)

(c - a)² = 0...................(3)

Now, Simplifying Eq. (1).

(a - b)² = 0

Transpose the square root. Remember, √0=0

a - b = 0

a = b ..............(4)

similarly, Simplifying Eq. (2), we get

(b - c)² = 0

Transpose the square root. Remember, √0=0

b - c = 0

b = c ..........(5)

again, Simplifying Eq. (3), we have

(c - a)² = 0

Transpose the square root. Remember, √0=0

c - a = 0

c = a .......(6)

From Equation No. (4), (5) & (6), we get:

a = b = c

HOPE THIS HELPS :D

Answer 2

Step-by-step explanation:

We have,

a² + b² + c² - ab - bc - ca = 0

=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 2 × 0

=> (a² - 2ab + b2) + (b² - 2bc + c²) + (c² - 2ac +

a²) = 0

=> (a-b)² + (b-c)² + (c-a)² = 0

=> a - b = 0, b - c = 0, c - a = 0

=> a = b, b = c and c = a => a = b = c

• Therefore, a = b = c