`|{:(a,b,c),(b,c,a),(c,a,b):}|` is equal to

Options

A. `a+b+c+abc`
B. `a^2+b^2+c^2+ab+bc+ca`
C. `3abc-a^2-b^3-c^3`
D. `a^3+b^3+c^3-3abc`

Answer

To find the value of the determinant \( D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \), we can expand it using the method of cofactor expansion along the first row.

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

2. **Expand Along the First Row**:
\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- For the first determinant:
\[ \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \]
- For the second determinant:
\[ \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \]
- For the third determinant:
\[ \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \]

4. **Substituting Back**:
\[ D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]

5. **Distributing**:
\[ D = acb - a^3 - b^3 + abc + abc - c^3 \]

6. **Combine Like Terms**:
\[ D = 3abc - (a^3 + b^3 + c^3) \]

### Final Answer:
\[ D = 3abc - (a^3 + b^3 + c^3) \]