If the value of the determinant $ \begin{vmatrix} a & 1& 1 \\ 1& b& 1\\ 1 & 1& c \end{vmatrix} $ is positive then
(a) abc > 1
(b) abc > -8
(c) abc < -8
(d) abc > -2
MY TRY
$ \begin{vmatrix} a & 1& 1 \\ 1& b& 1\\ 1 & 1& c \end{vmatrix} $ > 0
$\Rightarrow a(bc-1)-1(c-1)+1(1-b) > 0$ $\Rightarrow abc-a-b-c+2 >0$
Now i stuck at this step. Dont know how to solve this inequality in terms of abc??
Can anyone give me hint.
Take any $p\in\mathbb{R}$ and let $a=p$, $b=-1$ and $c=-1$. Then determinant is $$abc-a-b-c+2=p-p+1+1+2=4>0$$ and the product $abc$ is equal to $p$. Hence, the fact that the determinant is positive does not suffice to conclude anything about the value of $abc$.