Question

hello guys please solve this question.​

Answer 1

Step-by-step explanation:

a²+b²+c²-ab-bc-ca=0

1/2[2a²+2b²+2c²-2ab-2bc-2ca]=0

1/2[a²-2ab+b²+b²-2bc+c²+c²-2ca+a²]=0

1/2[(a-b)²+(b-c)²+(c-a)²]=0

(a-b)²+(b-c)²+(c-a)²=0

comparing both sides

(a-b)²=0

a=b

(b-c)²=0

b=c

(c-a)²=0

c=a

so,a=b=c