If `|(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2))|=k(a-b)(b-c)(c-a)`, then k is equal to

Options

A. `-2`
B. 1
C. 2
D. abc

Answer

To solve the determinant problem given by the equation

\[ |(1,a,a^2),(1,b,b^2),(1,c,c^2)| = k(a-b)(b-c)(c-a), \]

we need to find the value of \( k \).

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We start with the determinant:
\[ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \]

2. **Apply Row Operations**:
We can simplify the determinant by performing row operations. Let's subtract the second row from the first row and the third row from the second row:
\[ R_1 \rightarrow R_1 - R_2 \quad \text{and} \quad R_2 \rightarrow R_2 - R_3 \]
This gives us:
\[ D = \begin{vmatrix} 0 & a-b & a^2 - b^2 \\ 0 & b-c & b^2 - c^2 \\ 1 & c & c^2 \end{vmatrix} \]

3. **Factor the Differences**:
We can factor the differences in the second column:
\[ a^2 - b^2 = (a-b)(a+b) \quad \text{and} \quad b^2 - c^2 = (b-c)(b+c). \]
Thus, we rewrite the determinant:
\[ D = \begin{vmatrix} 0 & a-b & (a-b)(a+b) \\ 0 & b-c & (b-c)(b+c) \\ 1 & c & c^2 \end{vmatrix} \]

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

5. **Evaluate the Remaining Determinant**:
Now we can evaluate the remaining determinant:
\[ D' = \begin{vmatrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2 \end{vmatrix} \]
By expanding this determinant, we can eliminate the first column:
\[ D' = 0 - 0 + 1 \cdot \begin{vmatrix} 1 & a+b \\ c & c^2 \end{vmatrix} = c^2 - (a+b)c. \]

6. **Combine Everything**:
Thus, we have:
\[ D = (a-b)(b-c)(c^2 - (a+b)c). \]

7. **Compare with Given Expression**:
We know that:
\[ D = k(a-b)(b-c)(c-a). \]
By comparing both sides, we can see that:
\[ k = 1. \]

### Final Answer:
Thus, the value of \( k \) is:
\[ \boxed{1}. \]