both roots of the equation `(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0` are

Explanation

The given equation may be written as
`3x^(2)-2(a+b+c)x+(ab+bc+ac)=0`
`:."Discriminant"D=B^(2)-4AC`
`D=4(a+b+c)^(2)-4xx3(ab+bc+ac)`
`D=4(a^(2)+b^(2)+c^(2)+2ab+2bc+2ac)-12(ab+bc+ac)`
`=4(a^(2)+b^(2)+c^(2)-ab-bc-ac)`
`=2(2a^(2)+2b^(2)+2c^(2)-2ab-2bc-2ac)`
`=2[a^(2)+b^(2)-2ab+b^(2)+c^(2)-2ab+c^(2)+a^(2)-2ac]`
`=2[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]gt=0`
`[:'(a-b)^(2)gt=0,(b-c)^(2)gt==and(c-a)^(2)gt=0]`
Hence, both roots of the equation are real.
For equal root we must have D=0
`implies(a-b)^(2)+(b-c)^(2)+(c-a)^(2)=0`
`impliesa-b=0,b-c=0,c-a=0`
`impliesa=b,b=c,c=a`
`impliesa=b=c`
Hence, roots are equal only when a=b=c.