The value of `|(a,a^(2) - bc,1),(b,b^(2) - ca,1),(c,c^(2) - ab,1)|`, is

Options

A. 1
B. `-1`
C. 0
D. `-abc`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} a & a^2 - bc & 1 \\ b & b^2 - ca & 1 \\ c & c^2 - ab & 1 \end{vmatrix} \]

we will perform row operations to simplify the determinant.

### Step 1: Write the determinant
We start with the determinant as given:

\[ \Delta = \begin{vmatrix} a & a^2 - bc & 1 \\ b & b^2 - ca & 1 \\ c & c^2 - ab & 1 \end{vmatrix} \]

### Step 2: Perform row operations
We will perform the row operations \( R_2 \leftarrow R_2 - R_1 \) and \( R_3 \leftarrow R_3 - R_1 \):

\[ \Delta = \begin{vmatrix} a & a^2 - bc & 1 \\ b - a & (b^2 - ca) - (a^2 - bc) & 0 \\ c - a & (c^2 - ab) - (a^2 - bc) & 0 \end{vmatrix} \]

### Step 3: Simplify the second and third rows
Calculating the second row:

\[ (b^2 - ca) - (a^2 - bc) = b^2 - ca - a^2 + bc = b^2 - a^2 + bc - ca \]

Calculating the third row:

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

Now the determinant becomes:

\[ \Delta = \begin{vmatrix} a & a^2 - bc & 1 \\ b - a & b^2 - a^2 + bc - ca & 0 \\ c - a & c^2 - a^2 + bc - ab & 0 \end{vmatrix} \]

### Step 4: Expand the determinant
Since the last column has two zeros, we can expand along the last column:

\[ \Delta = 1 \cdot \begin{vmatrix} b - a & b^2 - a^2 + bc - ca \\ c - a & c^2 - a^2 + bc - ab \end{vmatrix} \]

### Step 5: Calculate the 2x2 determinant
Now we calculate the 2x2 determinant:

\[ \Delta = (b - a)(c^2 - a^2 + bc - ab) - (c - a)(b^2 - a^2 + bc - ca) \]

### Step 6: Factor and simplify
Notice that both terms can be factored and simplified. After simplification, we can find that:

\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \Delta = 0 \]