The value of `|(1,1,1),(bc,ca,ab),(b+c,c+a,a+b)|` is :

Options

A. 1
B. 0
C. `(a-b)(b-c)(c-a)`
D. `(a+b)(b+c)(c+a)`

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

we can use the properties of determinants to simplify our calculations.

### Step 1: Write the determinant

We start with the determinant:

\[ D = \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

### Step 2: Subtract the first row from the second and third rows

We can simplify the determinant by performing row operations. Specifically, we can subtract the first row from the second and third rows:

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

### Step 3: Expand the determinant

Now, we can expand the determinant along the first column:

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

### Step 4: Calculate the 2x2 determinant

Now we calculate the 2x2 determinant:

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

### Step 5: Simplify the expression

Now we simplify the expression:

\[ D = (c-b)(b-c) - (a-c)(a-b) \]

### Step 6: Factor the expression

We can factor this expression:

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

### Final Result

Thus, the value of the determinant is:

\[ D = (a-b)(b-c)(c-a) \]