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

Options

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

Explanation

To find the value of the determinant

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

we will perform a series of operations to simplify it.

### Step 1: Column Operation
We will perform the column operation \( C_1 \to C_1 + C_3 \). This means we will add the third column to the first column.

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

This simplifies to:

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

### Step 2: Factor Out Common Terms
Now, we can factor out \( (a + b + c) \) from the first column:

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

### Step 3: Row Operation
Next, we will perform the row operations \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):

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

This simplifies to:

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

### Step 4: Calculate the Determinant
Now, we can calculate the determinant of the 2x2 matrix:

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

Using the formula for the determinant of a 2x2 matrix:

\[ D = (a + b + c) \left( (2b - a - c)(c - a) - (b - a)(c + b - 2a) \right) \]

### Step 5: Expand and Simplify
Expanding the expression inside the determinant:

1. Expand \( (2b - a - c)(c - a) \)
2. Expand \( (b - a)(c + b - 2a) \)
3. Combine like terms.

After simplification, we will find that:

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

### Step 6: Final Result
The expression \( a^2 + b^2 + c^2 - ab - ac - bc \) can be recognized as \( \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) \).

Thus, the final value of the determinant is:

\[ D = (a + b + c) \cdot \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) \]