`|(bc,bc'+b'c,b'c'),(ca,ca'+c'a,c'a'),(ab,ab'+a'b,a'b')|` is equal to

Options

A. `(ab -a'b') (bc -b'c') (ca - c'a')`
B. `(ab + a'b') (bc + b'c') (ca + c'a')`
C. `(ab' - a'b) (bc' -b'c) (ca' - c'a)`
D. `(ab' + a'b) (bc' + b'c) (ca' + c'a)`

Answer

To evaluate the determinant given by

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

we will follow a systematic approach.

### Step 1: Factor out common terms from each row

We notice that each row has terms that can be factored out. We will factor out \(b'c'\) from the first row, \(c'a'\) from the second row, and \(a'b'\) from the third row.

\[ D = b'c'a'b' \begin{vmatrix} \frac{bc}{b'c'} & \frac{bc' + b'c}{b'c'} & \frac{b'c'}{b'c'} \\ \frac{ca}{c'a'} & \frac{ca' + c'a}{c'a'} & \frac{c'a'}{c'a'} \\ \frac{ab}{a'b'} & \frac{ab' + a'b}{a'b'} & \frac{a'b'}{a'b'} \end{vmatrix} \]

### Step 2: Simplify the determinant

After factoring out, we can simplify the determinant:

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

### Step 3: Perform row operations

Now, we can perform row operations to simplify the determinant further. We can subtract the first row from the second and third rows.

Row 2: \( R_2 \to R_2 - R_1 \)

Row 3: \( R_3 \to R_3 - R_1 \)

This gives us:

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

### Step 4: Expand the determinant

The determinant can now be expanded. Since the last column consists of zeros, the determinant simplifies to:

\[ D = b'c'a'b' \cdot 0 = 0 \]

### Final Result

Thus, the value of the determinant is:

\[ \boxed{0} \]