The correct option is C a3+b3+c3−3abc
△=∣∣
∣∣a−bb+cab−cc+abc−aa+bc∣∣
∣∣
Applying R1→R1+R2+R3, we get
△=(a+b+c)∣∣
∣∣021b−cc+abc−aa+bc∣∣
∣∣
Expanding along R1, we get
△=(a+b+c)[−2c(b−c)+2b(c−a)+(b−c)(a+b)−(c−a)(c+a)]
=(a+b+c)(a2+b2+c2−ab−bc−ac)
=a3+b3+c3−3abc