Doubt by Bhavya
Solution :
a²+b²+c²-ab-bc-ca=0
Multiply both sides by 2
Multiply both sides by 2
2(a²+b²+c²-ab-bc-ca)=2×0
2a²+2b²+2c²-2ab-2bc-2ca=0
a²+b²-2ab+b²+c²-2bc+c²+a²-2ca=0
(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)=0
(a-b)²+(b-c)²+(c-a)²=0
As we know, the square of a number can never be negative so the LHS of the above statement will only be zero if and and if (a-b), (b-c) and (c-a) all individually are equal to zero.
(a-b)²+(b-c)²+(c-a)²=0
As we know, the square of a number can never be negative so the LHS of the above statement will only be zero if and and if (a-b), (b-c) and (c-a) all individually are equal to zero.
i.e.
a-b=0 or a=b — (1)
b-c=0 or b=c — (2)
c-a=0 or c=a — (3)
from equations (1), (2) and (3)
a=b=c
b-c=0 or b=c — (2)
c-a=0 or c=a — (3)
from equations (1), (2) and (3)
a=b=c
Hence Proved.