`|(b+c,a,a),(b,c+a,b),(c,c,a+b)|=`

Options

A. abc
B. 2abc
C. 3abc
D. 4abc

Answer

To solve the determinant \( D = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \), we will use the method of expansion along the first row.

### Step 1: Expand the Determinant
Using the first row for expansion, we have:

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

### Step 2: Calculate the First 2x2 Determinant
Now, we calculate the first 2x2 determinant:

\[ \begin{vmatrix} c+a & b \\ c & a+b \end{vmatrix} = (c+a)(a+b) - bc \]
Expanding this gives:
\[ = ac + ab + a^2 + bc - bc = ac + ab + a^2 \]

### Step 3: Calculate the Second 2x2 Determinant
Next, we calculate the second 2x2 determinant:

\[ \begin{vmatrix} b & b \\ c & a+b \end{vmatrix} = b(a+b) - bc = ab + b^2 - bc \]

### Step 4: Calculate the Third 2x2 Determinant
Now, we calculate the third 2x2 determinant:

\[ \begin{vmatrix} b & c+a \\ c & c \end{vmatrix} = b(c) - c(c+a) = bc - (c^2 + ac) \]

### Step 5: Substitute Back into the Determinant
Now we substitute these values back into our expression for \( D \):

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

### Step 6: Simplify the Expression
Now we simplify the expression step by step:

1. **Expand the first term**:
\[ (b+c)(ac + ab + a^2) = abc + ab^2 + a^2b + ac^2 + a^2c + abc \]
This simplifies to:
\[ 2abc + ab^2 + a^2b + ac^2 + a^2c \]

2. **Expand the second term**:
\[ -a(ab + b^2 - bc) = -a^2b - ab^2 + abc \]

3. **Expand the third term**:
\[ a(bc - c^2 - ac) = abc - ac^2 - a^2c \]

### Step 7: Combine All Terms
Combining all the terms gives:
\[ D = 2abc + ab^2 + a^2b + ac^2 + a^2c - a^2b - ab^2 + abc + abc - ac^2 - a^2c \]

### Step 8: Final Simplification
Now, we combine like terms:
- The \( abc \) terms: \( 2abc + abc + abc = 4abc \)
- The \( ab^2 \) terms: \( ab^2 - ab^2 = 0 \)
- The \( a^2b \) terms: \( a^2b - a^2b = 0 \)
- The \( ac^2 \) terms: \( ac^2 - ac^2 = 0 \)
- The \( a^2c \) terms: \( a^2c - a^2c = 0 \)

Thus, we find that:
\[ D = 4abc \]

### Final Answer
The value of the determinant is \( \boxed{4abc} \).