The value of `|(a-b,b+c,a),(b-a,c+a,b),(c-a,a+b,c)|` is

Options

A. `a^(3)+b^(3)+c^(3)`
B. `3bc`
C. `a^(3)+b^(3)+c^(3)-3abc`
D. None of these

Explanation

To find the value of the determinant

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

we will perform a series of transformations to simplify the determinant.

### Step 1: Column Transformation
We will perform the column operation \( C_2 \rightarrow C_2 + C_3 \). This means we will add the third column to the second column.

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

This simplifies to:

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

### Step 2: Factor Out Common Terms
Notice that the second column can be factored out as \( A + B + C \):

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

### Step 3: Row Transformation
Now, we will perform row operations to simplify the determinant further. We will perform \( R_1 \rightarrow R_1 - R_2 \) and \( R_2 \rightarrow R_2 - R_3 \):

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

This simplifies to:

\[ D = (a+b+c) \begin{vmatrix} 2(a-b) & 0 & a-b \\ b-c & 0 & b-c \\ c-a & 1 & c \end{vmatrix} \]

### Step 4: Factor Out Common Terms Again
From the first row, we can factor out \( 2(a-b) \) and from the second row, we can factor out \( (b-c) \):

\[ D = (a+b+c)(a-b)(b-c) \begin{vmatrix} 1 & 0 & 1 \\ 1 & 0 & 1 \\ c-a & 1 & c \end{vmatrix} \]

### Step 5: Evaluate the Remaining Determinant
The determinant now has two identical rows, which means it evaluates to zero:

\[ D = (a+b+c)(a-b)(b-c) \cdot 0 = 0 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{0} \]