[tex](x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a) = 3x^2 - 2(a+b+c)x+(ab+bc+ac)[/tex]
which has equal roots if the discriminant is zero, i.e.
[tex]4(a+b+c)^2-12(ab+bc+ac) = 4a^2+4b^2+4c^2-4ab-4bc-4ac = 2(a-b)^2+2(b-c)^2+2(c-a)^2 = 0[/tex]
The only way the sum of squares is zero is if all the squares are zero, which implies [tex]a=b=c[/tex].