If Δ=−xabb−xaab−x then a factor of ∆ is
a+b+x
x2−(a−b)x+a2+b2+ab
x2+(a+b)x+a2+b2−ab
a+b-x
Applying C1→C1+C2+C3 we get
Δ=a+b−xaba+b−x−xaa+b−xb−x=(a+b−x)1ab1−xa1b−x=(a+b−x)1ab0−x−aa−b0b−a−x−b
[Applying R2→R2−R1 and R3→R3−R1]
=(a+b−x)(x+a)(x+b)+(a−b)2 [expanding along C1=(a+b−x)x2+(a+b)x+a2+b2−ab