(x−a)(x−b)+(x−b)(x−c)+(x−c)(x−a)=0=3x2−2(a+b+c)x+(ab+bc+ca)=0∵ Roots are equal ∴B2−4AC=04(a+b+c)2−4(3)(ab+bc+ca)=0 ⇒ 4{(a+b+c)2−3(ab+bc+ca)}=0 ⇒ 4{a2+b2+c2−ab−bc−ca}=0 ⇒ 2{(a−b)2+(b−c)2+(c−a)2}=0 ⇒ (a−b)2+(b−c)2+(c−a)2=0 Hence, two roots are always equal a=b or b=c or c=a.