Doubt by Yathartha
Solution :
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
x(x-b)-a(x-b)+x(x-c)-b(x-c)+x(x-a)-c(x-a)=0
x²-bx-ax+ab+x²-cx-bx+bc+x²-ax-cx+ca=0
x(x-b)-a(x-b)+x(x-c)-b(x-c)+x(x-a)-c(x-a)=0
x²-bx-ax+ab+x²-cx-bx+bc+x²-ax-cx+ca=0
3x²-2ax-2bx-2cx-2dx+ab+bc+ca=0
3x²-2(a+b+c)x+(ab+bc+ca)=0
Now, comparing the coefficients
A = 3
B = -2(a+b+c)
C = (ab+bc+ca)
B = -2(a+b+c)
C = (ab+bc+ca)
D = B²-4AC
D = [-2(a+b+c)]²-4(3)(ab+bc+ca)
But roots are real and equal.
So D = 0
0 = 4(a+b+c)²-12(ab+bc+ca)
0 = 4(a²+b²+c²+2ab+2bc+2ca)-12(ab-bc-ca)
D = [-2(a+b+c)]²-4(3)(ab+bc+ca)
But roots are real and equal.
So D = 0
0 = 4(a+b+c)²-12(ab+bc+ca)
0 = 4(a²+b²+c²+2ab+2bc+2ca)-12(ab-bc-ca)
0 = 4a²+4b²+4c²+8ab+8bc+8ca-12ab-12bc-12ca
0 = 4a²+4b²+4c²-4ab-4bc-4ca
0 = 2[2a²+2b²+2c²-2ab-2bc-2ca]
0/2 = 2a²+2b²+2c²-2ab-2bc-2ca
0 = 2a²+2b²+2c²-2ab-2bc-2ca
0 = a²+a²+b²+b²+c²+c²-2ab-2bc-2ca
0/2 = 2a²+2b²+2c²-2ab-2bc-2ca
0 = 2a²+2b²+2c²-2ab-2bc-2ca
0 = a²+a²+b²+b²+c²+c²-2ab-2bc-2ca
0 = (a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)
0 = (a-b)²+(b-c)²+(c-a)²
The above statement is only possible when
a-b=0, b-c=0, c-a=0
a=b, b=c, c=a
⇒a=b=c
0 = (a-b)²+(b-c)²+(c-a)²
The above statement is only possible when
a-b=0, b-c=0, c-a=0
a=b, b=c, c=a
⇒a=b=c
Hence Proved.