If a, b and c are real numbers then the roots of the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 are always

Step-by-step solution

Given equation is (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 3 x²-2(b+a+c) x+a b+b c+c a=0 Now, here A=3, B=-2(a+b+c)C=a b+b c+c a D= B²-4 A C = (-2(a+b+c))²-4(3)(a b+b c+c a) = 4(a+b+c)²-12(a b+b c+c a) =2 a²+b²+c²-a b-b c-c a =2 1 2 (a-b)²-(b-c)²+(c-a)² 0