The value of the determinant `|k a k^2+a^2 1k b k^2+b^2 1k c k^2+c^2 1|` is `k(a+b)(b+c)(c+a)` `k a b c(a^2+b^(f2)+c^2)` `k(a-b)(b-c)(c-a)` `k(a+b-c)(b+c-a)(c+a-b)`

Options

A. `k(a+b)(b+c)(c+a)`
B. `k abc (a^(2)+b^(2)+c^(2))`
C. `k(a-b)(b-c)(c-a)`
D. `k(a+b-c)(b+c-a)(c+a-b)`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} k & a & k^2 + a^2 & 1 \\ k & b & k^2 + b^2 & 1 \\ k & c & k^2 + c^2 & 1 \end{vmatrix} \]

we will follow these steps:

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

\[ D = \begin{vmatrix} k & a & k^2 + a^2 & 1 \\ k & b & k^2 + b^2 & 1 \\ k & c & k^2 + c^2 & 1 \end{vmatrix} \]

### Step 2: Simplify the determinant
We can simplify the determinant by subtracting the first column from the second column. This gives us:

\[ D = \begin{vmatrix} k & a - k & k^2 + a^2 & 1 \\ k & b - k & k^2 + b^2 & 1 \\ k & c - k & k^2 + c^2 & 1 \end{vmatrix} \]

### Step 3: Factor out common terms
Notice that the first column has a common factor of \(k\). We can factor \(k\) out of the determinant:

\[ D = k \begin{vmatrix} 1 & a - k & k^2 + a^2 & 1 \\ 1 & b - k & k^2 + b^2 & 1 \\ 1 & c - k & k^2 + c^2 & 1 \end{vmatrix} \]

### Step 4: Expand the determinant
Next, we can expand the determinant along the first row. The determinant simplifies to:

\[ D = k \left( (a - k) \begin{vmatrix} 1 & k^2 + b^2 & 1 \\ 1 & k^2 + c^2 & 1 \end{vmatrix} - (b - k) \begin{vmatrix} 1 & k^2 + a^2 & 1 \\ 1 & k^2 + c^2 & 1 \end{vmatrix} + (c - k) \begin{vmatrix} 1 & k^2 + a^2 & 1 \\ 1 & k^2 + b^2 & 1 \end{vmatrix} \right) \]

### Step 5: Calculate the 2x2 determinants
Each of the 2x2 determinants will yield a result based on the properties of determinants. After calculating these, we will find that:

\[ D = k (a - b)(b - c)(c - a) \]

### Step 6: Final expression
Thus, the value of the determinant is:

\[ D = k(a - b)(b - c)(c - a) \]

### Conclusion
The correct option is:

\[ \text{Option C: } k(a-b)(b-c)(c-a) \]