The correct option is C abc>−8
Δ=∣∣
∣∣0011−ab−111−ac1−cc∣∣
∣∣[C1→C1−aC3;C2→C2−C3]=(1−a)(1−c)−(b−1)(1−ac)=1−a−c+ac−b+abc+1−ac=(2+abc)−(a+b+c),as Det>0i,e.abc+2>a+b+c{∵a+b+c>3(abc)1/3}abc+2>3(abc)1/3⇒x3−3x+2>0,(Letx=(abc)13)⇒(x−1)2(x+2)>0i.e.x>−2i.e.x3=abc>−8