The correct option is C (1+a2+b2)3
Δ=1ab∣∣
∣
∣∣b(1+a2−b2)2ab2−2b22a2ba(1−a2+b2)2a22b−2a1−a2−b2∣∣
∣
∣∣by(R1×b,R2×a)∣∣
∣
∣∣1+a2−b22b2−2b22a21−a2+b22a22−21−a2−b2∣∣
∣
∣∣by(C1b,C2a)∣∣
∣
∣∣1+a2+b20−2b21+a2+b21+a2+b22a20−(1+a2+b2)1−a2−b2∣∣
∣
∣∣(C1→C1+C2,C2→C2+C3)=(1+a2+b2)3