If the value of the determinant `|(a,1,1),(1,b,1),(1,1,c)|` is positive, where `a ne be ne c`, then the value of abc.

Options

A. Cannot be less than 1
B. Is greater than -8
C. Is less than -8
D. Must be greater than 8

Answer

To solve the problem, we need to evaluate the determinant \( D = \begin{vmatrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{vmatrix} \) and find the conditions under which this determinant is positive, given that \( a \neq b \neq c \).

### Step 1: Calculate the Determinant

We will use the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ D = a \begin{vmatrix} b & 1 \\ 1 & c \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & c \end{vmatrix} + 1 \begin{vmatrix} 1 & b \\ 1 & 1 \end{vmatrix} \]

Calculating the \( 2 \times 2 \) 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:

\[ D = a(bc - 1) - (c - 1) + (1 - b) \]

Simplifying this expression:

\[ D = abc - a - c + 1 + 1 - b \]
\[ D = abc - a - b - c + 2 \]

### Step 2: Set the Determinant Greater than Zero

According to the problem, we need \( D > 0 \):

\[ abc - a - b - c + 2 > 0 \]

This can be rearranged to:

\[ abc > a + b + c - 2 \]

### Step 3: Apply the AM-GM Inequality

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we know that:

\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]

This implies:

\[ a + b + c \geq 3\sqrt[3]{abc} \]

### Step 4: Compare the Two Inequalities

From our previous inequality \( abc > a + b + c - 2 \), we can substitute the AM-GM result:

\[ abc > 3\sqrt[3]{abc} - 2 \]

Let \( x = \sqrt[3]{abc} \). Then \( abc = x^3 \), and we can rewrite the inequality as:

\[ x^3 > 3x - 2 \]

### Step 5: Solve the Inequality

Rearranging gives us:

\[ x^3 - 3x + 2 > 0 \]

To find the roots of the equation \( x^3 - 3x + 2 = 0 \), we can use synthetic division or trial and error. Testing \( x = 1 \):

\[ 1^3 - 3(1) + 2 = 0 \]

Thus, \( x - 1 \) is a factor. We can factor the polynomial:

\[ x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) \]

Factoring \( x^2 + x - 2 \) gives:

\[ (x - 1)(x + 2)(x - 1) = 0 \]

The roots are \( x = 1 \) and \( x = -2 \). The cubic polynomial changes sign at these points.

### Step 6: Analyze the Intervals

We analyze the intervals:

1. For \( x < -2 \): \( (x - 1)(x + 2) < 0 \)
2. For \( -2 < x < 1 \): \( (x - 1)(x + 2) > 0 \)
3. For \( x > 1 \): \( (x - 1)(x + 2) > 0 \)

Thus, the inequality \( x^3 - 3x + 2 > 0 \) holds for \( x > 1 \) or \( x < -2 \).

### Step 7: Conclusion

Since \( x = \sqrt[3]{abc} \), we have:

1. \( \sqrt[3]{abc} > 1 \) implies \( abc > 1^3 = 1 \)
2. \( \sqrt[3]{abc} < -2 \) implies \( abc < (-2)^3 = -8 \)

However, since \( a, b, c \) must be distinct and positive for the determinant to be positive, we conclude:

\[ abc > 1 \]

### Final Answer

Thus, the value of \( abc \) is greater than 1.