Choose the correct answer from the following :
The value of `|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|is:`

Options

A. `(a-b)(b-c)(c-a)`
B. `-(a-b)(b-c)(c-a)`
C. `(a+b+c)(a-b)(b-c)(c-a)`
D. None of these

Answer

To solve the determinant \( | \begin{pmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix} | \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant using row operations. We will perform the following operations:
- \( R_1 \rightarrow R_1 - R_2 \)
- \( R_2 \rightarrow R_2 - R_3 \)

This gives us:
\[ D = \begin{vmatrix} 0 & a-b & a^2 - b^2 \\ 0 & b-c & b^2 - c^2 \\ 1 & c & c^2 \end{vmatrix} \]

### Step 3: Factor the Differences of Squares
Recall that \( a^2 - b^2 = (a-b)(a+b) \) and \( b^2 - c^2 = (b-c)(b+c) \). We can rewrite the determinant as:
\[ D = \begin{vmatrix} 0 & a-b & (a-b)(a+b) \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & c^2 \end{vmatrix} \]

### Step 4: Factor Out Common Terms
We can factor out \( (a-b) \) from the first row and \( (b-c) \) from the second row:
\[ D = (a-b)(b-c) \begin{vmatrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2 \end{vmatrix} \]

### Step 5: Expand the Determinant
Now we can expand the determinant:
\[ D = (a-b)(b-c) \begin{vmatrix} 1 & a+b \\ 1 & b+c \\ c & c^2 \end{vmatrix} \]

### Step 6: Calculate the 2x2 Determinant
Calculating the 2x2 determinant:
\[ = 1 \cdot (b+c) - 1 \cdot (a+b) = (b+c) - (a+b) = c - a \]

### Step 7: Final Expression for the Determinant
Thus, we have:
\[ D = (a-b)(b-c)(c-a) \]

### Conclusion
The value of the determinant \( | \begin{pmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{pmatrix} | \) is:
\[ D = (a-b)(b-c)(c-a) \]