 If the value of the determinant $\left| {\begin{array}{*{20}{c}} a&1&1 \\ 1&b&1 \\ 1&1&c \end{array}} \right|$ is positive, then$\left( a \right)abc > 1$$\left( b \right)abc > - 8$$\left( c \right)abc < - 8$$\left( d \right)abc > - 2$ Verified
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Hint: In this particular question use the concept that if something is positive than it is always greater than zero, later on in the solution use the concept that arithmetic mean is always greater than or equal to geometric mean so use these concepts to reach the solution of the question.

Given determinant
$\left| {\begin{array}{*{20}{c}} a&1&1 \\ 1&b&1 \\ 1&1&c \end{array}} \right|$
Now we have to find out the value of determinant if the determinant is positive.
So as the determinant is positive so the value of determinant is always greater than zero.
$\Rightarrow \left| {\begin{array}{*{20}{c}} a&1&1 \\ 1&b&1 \\ 1&1&c \end{array}} \right| > 0$
Now expand the determinant we have,
$\Rightarrow \left[ {a\left| {\begin{array}{*{20}{c}} b&1 \\ 1&c \end{array}} \right| - 1\left| {\begin{array}{*{20}{c}} 1&1 \\ 1&c \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} 1&b \\ 1&1 \end{array}} \right|} \right] > 0$
Now expand the mini determinants we have,
$\Rightarrow \left[ {a\left( {bc - 1} \right) - 1\left( {c - 1} \right) + 1\left( {1 - b} \right)} \right] > 0$
$\Rightarrow \left[ {a\left( {bc - 1} \right) - 1\left( {c - 1} \right) + 1\left( {1 - b} \right)} \right] > 0$
Now simplify we have,
$\Rightarrow \left[ {abc - a - c + 1 + 1 - b} \right] > 0$
$\Rightarrow abc - a - b - c + 2 > 0$
$\Rightarrow abc + 2 > a + b + c$................. (1)
Now consider three numbers a, b, and c.
Arithmetic mean (A.M) of the numbers is $= \dfrac{{a + b + c}}{3}$
And the Geometric mean (G.M) of the numbers is $\sqrt{{abc}}$
Now as we know that
$A.M \geqslant G.M$, so use this property
$\therefore \dfrac{{a + b + c}}{3} \geqslant \sqrt{{abc}}$
$\Rightarrow a + b + c \geqslant 3\sqrt{{abc}}$............... (2)
Now from equation (1) and (2) we have,
$\Rightarrow abc + 2 > 3\sqrt{{abc}}$
Now let abc = ${x^3}$.................. (3), so substitute this in the above equation we have,
$\Rightarrow {x^3} + 2 > 3x$
$\Rightarrow {x^3} - 3x + 2 > 0$................... (4)
Now first solve the cubic equation, ${x^3} - 3x + 2$
$\Rightarrow {x^3} - 3x + 2 = 0$
So solve this by hit and trial method, so substitute x = 1 in the above equation we have,
$\Rightarrow {1^3} - 3 + 2 = 0$
$\Rightarrow 0 = 0$
So 1 is the root of the equation, so (x - 1) is the factor of the cubic equation.
Now substitute x = -2 in the above equation we have,
$\Rightarrow {\left( { - 2} \right)^3} - 3\left( { - 2} \right) + 2 = 0$
$\Rightarrow - 8 + 6 + 2 = 0$
$\Rightarrow 0 = 0$
So -2 is the root of the equation, so (x + 2) is the factor of the cubic equation.
Let the other factor be (x - a).
Therefore, (x - a) (x + 2) (x - 1) = ${x^3} - 3x + 2$
$\Rightarrow \left( {x - a} \right)\left( {{x^2} + x - 2} \right) = {x^3} - 3x + 2$
$\Rightarrow \left( {{x^3} + \left( {1 - a} \right){x^2} - \left( {2 + a} \right)x + 2a} \right) = {x^3} - 3x + 2$
Now on comparing the constant term we have,
$\Rightarrow 2a = 2$
Therefore, a = 1
So from equation (4) we have,
$\Rightarrow {\left( {x - 1} \right)^2}\left( {x + 2} \right) > 0$
Now as we know that the square term is always positive for any value of x.
But (x + 2) cannot be positive for any value of x.
So for the determinant to be positive (x + 2) must be greater than zero.
Therefore, x > - 2
Now take cube on both sides we have,
$\Rightarrow {x^3} > {\left( { - 2} \right)^3}$
$\Rightarrow {x^3} > - 8$
Now from equation (3) we have,
$\Rightarrow abc > - 8$

Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the arithmetic mean of three numbers (a, b, and c) is $\dfrac{{a + b + c}}{3}$ and the geometric mean of three numbers (a, b, and c) is $\sqrt{{abc}}$, and always recall that the quickly solution of a cubic equation if the solution is an integer value is by hit and trial method as applied above.