A) \[abc>1\]
B) \[abc>-8\]
C) \[abc<-8\]
D) \[abc>-2\]
Correct Answer: B
Solution :
Let \[\Delta =\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \\ \end{matrix} \right|=abc-(a+b+c)+2\] \[\because \] \[\Delta >0\Rightarrow abc+2>a+b+c\] \[\Rightarrow \] \[abc+2>3{{(abc)}^{1/3}}\] \[\left[ \because AM>GM\Rightarrow \frac{a+b+c}{3}>{{(abc)}^{1/3}} \right]\] \[\Rightarrow \] \[{{x}^{2}}+2>3x,\]where \[x={{(abc)}^{1/3}}\] \[\Rightarrow \] \[{{x}^{3}}-3x+2>0\]\[\Rightarrow \] \[{{(x-1)}^{2}}(x+2)>0\] \[\Rightarrow \] \[x+2>0\Rightarrow x>-2\Rightarrow {{(abc)}^{1/3}}>-2\] \[\Rightarrow \] \[abc>-8\]
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