A root of the equation Δ=0x−ax−bx+a0x−cx+bx+c0=0 is

When we substitute x = 0, ∆ becomes      0−a−ba0−cbc0which is equal to 0 as ∆ is skew symmetric determinant of odd order.Alternative Solution Evaluating ∆ along R1, we get−(x−a)x+ax−cx+b0+(x−b)x+a0x+bx+c=0⇒(x−a)(x+b)(x−c)+(x−b)(x+a)(x+c)=0⇒x3+(−a+b−c)x2+(−ab+ac−bc)x+abc+x3+(a−b+c)x2+(−ab+ac−bc)−abc=0⇒2x3−2(ab−ac+bc)x=0⇒2xx2−(ab−ac+bc)=0 ⇒x=0 or x=±ab−ac+bc∴ one of the roots of the equation is 0.