if the value of the determinant `|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}|` is positivie then
`(a,b,c lt 0)`

Options

A. `abc gt 1`
B. ` abc gt -8`
C. ` abc gt -8`
D. `abc gt -2`

Explanation

To solve the determinant problem, we will follow these steps:

### Step 1: Write down the determinant
We are given the determinant:
\[ D = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} \]

### Step 2: Calculate the determinant
Using the formula for the determinant of a 3x3 matrix, we have:
\[ D = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} = bc - 1\)
2. \(\begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} = c - 1\)
3. \(\begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} = 1 - b\)

Substituting these back into the determinant:
\[ D = a(bc - 1) - (c - 1) + (1 - b) \]
\[ D = abc - a - c + 1 + 1 - b \]
\[ D = abc - a - b - c + 2 \]

### Step 3: Set the determinant greater than zero
We know from the problem statement that:
\[ abc - a - b - c + 2 > 0 \]
Rearranging gives:
\[ abc + 2 > a + b + c \]

### Step 4: Apply the AM-GM inequality
Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
This implies:
\[ a + b + c \geq 3\sqrt[3]{abc} \]

### Step 5: Substitute back into the inequality
Substituting this into our earlier inequality:
\[ abc + 2 > 3\sqrt[3]{abc} \]
Let \(x = \sqrt[3]{abc}\), then we can rewrite the inequality:
\[ x^3 + 2 > 3x \]
Rearranging gives:
\[ x^3 - 3x + 2 > 0 \]

### Step 6: Factor the cubic polynomial
Factoring \(x^3 - 3x + 2\):
\[ (x - 1)^2(x + 2) > 0 \]

### Step 7: Analyze the factors
The expression \((x - 1)^2(x + 2) > 0\) is positive when:
1. \(x - 1 = 0\) gives \(x = 1\) (double root, does not change sign)
2. \(x + 2 = 0\) gives \(x = -2\)

The critical points are \(x = -2\) and \(x = 1\). The expression is positive for:
- \(x < -2\) (not possible since \(x = \sqrt[3]{abc}\) must be real)
- \(-2 < x < 1\) (valid)
- \(x > 1\) (valid)

### Step 8: Conclusion
Since \(x = \sqrt[3]{abc}\), we have:
\[ \sqrt[3]{abc} > -2 \Rightarrow abc > (-2)^3 = -8 \]

Thus, the final answer is:
\[ abc > -8 \]