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

The given equation is (x−b)(x−c)+(x−a)(x−c)+(x−a)(x−b)=0 ⇒[x2−(b+c)x+bc]+[x2−(a+c)x+ac]+[x2−(a+b)x+ab]=0⇒3x2−2(a+b+c)x+(bc+ca+ab)=0Here A=3,B=−2(a+b+c),C=(bc+ca+ab)∴  dicriminant B2−4AC=4(a+b+c)2−12(bc+ca+ab)                                             =4(a2+b2+c2+2ab+2bc+2ac)−12(bc+ca+ab)                                     =2[(a−b)2+(b−c)2+(c−a)2]≥0           ∴    Both the roots are real.