What is the value of the determinant `|(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b))|` ?

Options

A. 0
B. abc
C. ab + bc + ca
D. abc (a + b + c)

Answer

To find the value of the determinant

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

we can apply row operations to simplify the determinant.

### Step 1: Subtract Row 2 from Row 1 and Row 3 from Row 2

Let's perform the following row operations:

- \( R_1 \leftarrow R_1 - R_2 \)
- \( R_3 \leftarrow R_3 - R_2 \)

This gives us:

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

### Step 2: Simplify the elements

Calculating the elements in the first row:

- The first element is \( 0 \).
- The second element is \( bc - ca \).
- The third element simplifies as follows:
\[ a(b+c) - b(c+a) = ab + ac - bc - ab = ac - bc = c(a - b) \]

Now the determinant looks like:

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

### Step 3: Expand the determinant

Since the first column has two zeros, we can expand along the first column:

\[ D = 0 \cdot \text{(minor)} - 0 \cdot \text{(minor)} + 1 \cdot \begin{vmatrix} bc - ca & c(a - b) \\ ab - ca & c(a+b) - b(c+a) \end{vmatrix} \]

This simplifies to:

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

### Step 4: Calculate the 2x2 determinant

Now we calculate the 2x2 determinant:

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

### Step 5: Simplify the expression

Calculating \( c(a+b) - b(c+a) \):

\[ c(a+b) - b(c+a) = ca + cb - bc - ab = ca - ab \]

Now substituting back into the determinant:

\[ D = (bc - ca)(ca - ab) - c(a - b)(ab - ca) \]

### Step 6: Factor out common terms

Notice that both terms have common factors. After simplifying, we find that:

\[ D = 0 \]

### Final Answer

Thus, the value of the determinant is

\[ \boxed{0} \]