Both the roots of the given equation(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always

Given equation(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0can be re-written as3x2-2(a+b+c)x+(ab+bc+ca)=0 ∆=4{(a+b+c)2-3(ab+bc+ca)}  (∵b2-4ac=∆)  =4(a2+b2+c2-ab-bc-ac) =2{(a-b)2+(b-c)2+(c-a)2}≥0Hence both roots are always real.