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