If `a ne b ne c` are all positive, then the value of the determinant `|{:(a,b,c),(b,c,a),(c,a,b):}|`is

Options

A. `ge 0`
B. `gt 0`
C. `le -1`
D. `lt 0`

Answer

To find the value of the determinant

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

where \( a \neq b \neq c \) and all are positive, we will follow these steps:

### Step 1: Write the determinant
We start by writing the determinant explicitly:

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

### Step 2: Expand the determinant
Using the rule of Sarrus or cofactor expansion, we can expand the determinant:

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

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2\)

2. \(\begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac\)

3. \(\begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2\)

Substituting these back into the determinant:

\[ D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]

### Step 3: Simplify the expression
Expanding this gives:

\[ D = acb - a^3 - b^3 + abc + abc - c^3 \]

Combining like terms:

\[ D = 3abc - (a^3 + b^3 + c^3) \]

### Step 4: Factor the expression
We can use the identity for the sum of cubes:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

Thus, we can rewrite \(D\) as:

\[ D = -(a^3 + b^3 + c^3 - 3abc) = -\left((a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc)\right) \]

### Step 5: Analyze the signs
Since \(a\), \(b\), and \(c\) are all positive and distinct, both \(a + b + c\) and \(a^2 + b^2 + c^2 - ab - ac - bc\) are positive. Therefore, the product is positive, and since we have a negative sign in front, we conclude:

\[ D < 0 \]

### Conclusion
Thus, the value of the determinant is negative:

\[ D < 0 \]