The correct options are
B x2+(a+b)x+a2+b2−ab
C a+b−x
Applying C1→C1+C2+C3
Δ=∣∣
∣∣a+b−xaba+b−x−xaa+b−xb−x∣∣
∣∣=(a+b−x)∣∣
∣∣1ab1−xa1b−x∣∣
∣∣
Again applying R2→R2−R1,R3→R3−R1
Δ=(a+b−x)∣∣
∣∣1ab0−x−aa−b0b−a−x−b∣∣
∣∣=(a+b−x)((x+a)(x+b)+(a−b)2)=(a+b−x)(x2+(a+b)x+a2+b2−ab)