`|(a+b,a,b),(a,a+c,c),(b,c,b+c)|` is equal to

Options

A. 4abc
B. abc
C. `a^(2)b^(2)c^(2)`
D. `4a^(2)bc`

Answer

To find the value of the determinant \( |(a+b,a,b),(a,a+c,c),(b,c,b+c)| \), we will follow the steps outlined below:

### Step 1: Write the Determinant
We start by writing the determinant in a standard form:

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

### Step 2: Expand the Determinant
We will expand the determinant using the first row. The formula for the determinant of a 3x3 matrix is:

\[ D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values from our matrix:

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

### Step 3: Calculate the 2x2 Determinants
Now we will calculate each of the 2x2 determinants.

1. For the first determinant:
\[ \begin{vmatrix} a+c & c \\ c & b+c \end{vmatrix} = (a+c)(b+c) - c^2 = ab + ac + bc + c^2 - c^2 = ab + ac + bc \]

2. For the second determinant:
\[ \begin{vmatrix} a & c \\ b & b+c \end{vmatrix} = a(b+c) - bc = ab + ac - bc \]

3. For the third determinant:
\[ \begin{vmatrix} a & a+c \\ b & c \end{vmatrix} = a(c) - b(a+c) = ac - ab - bc \]

### Step 4: Substitute Back into the Determinant
Now we substitute these results back into the expression for \( D \):

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

### Step 5: Expand and Simplify
Now we will expand and simplify:

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

2. The second term:
\[ - a(ab + ac - bc) = -a^2b - a^2c + abc \]

3. The third term:
\[ b(ac - ab - bc) = abc - ab^2 - b^2c \]

### Step 6: Combine All Terms
Now we combine all the terms:

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

Combining like terms, we find:

- The \( a^2b \) and \( -a^2b \) cancel.
- The \( a^2c \) and \( -a^2c \) cancel.
- The \( 2abc + abc + abc = 4abc \).
- The \( ab^2 - ab^2 = 0 \).
- The \( b^2c - b^2c = 0 \).

Thus, we have:

\[ D = 4abc \]

### Final Answer
The value of the determinant is:

\[ \boxed{4abc} \]