What is the value of k, if `|(k,b+c,b^(2)+c^(2)),(k,c+a,c^(2) + a^(2)),(k,a+b,a^(2)+b^(2))| = (a - b) (b - c) (c - a)` ?

Options

A. A. 1
B. B. -1
C. C. 2
D. D. 0

Answer

To find the value of \( k \) in the determinant equation

\[ \left| \begin{array}{ccc} k & b+c & b^2+c^2 \\ k & c+a & c^2+a^2 \\ k & a+b & a^2+b^2 \end{array} \right| = (a-b)(b-c)(c-a), \]

we will simplify the determinant step by step.

### Step 1: Simplify the Determinant
We start with the determinant:

\[ D = \left| \begin{array}{ccc} k & b+c & b^2+c^2 \\ k & c+a & c^2+a^2 \\ k & a+b & a^2+b^2 \end{array} \right|. \]

Since the first column has \( k \) in all rows, we can perform row operations to simplify our determinant. Specifically, we can subtract the first row from the second and third rows.

### Step 2: Perform Row Operations
Perform the following operations:

- \( R_2 \leftarrow R_2 - R_1 \)
- \( R_3 \leftarrow R_3 - R_1 \)

This gives us:

\[ D = \left| \begin{array}{ccc} k & b+c & b^2+c^2 \\ 0 & (c+a) - (b+c) & (c^2+a^2) - (b^2+c^2) \\ 0 & (a+b) - (b+c) & (a^2+b^2) - (b^2+c^2) \end{array} \right|. \]

### Step 3: Simplify the Rows
Now, simplifying the second and third rows:

- For \( R_2 \):
\[ (c+a) - (b+c) = a - b, \]
\[ (c^2 + a^2) - (b^2 + c^2) = a^2 - b^2. \]

- For \( R_3 \):
\[ (a+b) - (b+c) = a - c, \]
\[ (a^2 + b^2) - (b^2 + c^2) = a^2 - c^2. \]

Thus, the determinant becomes:

\[ D = \left| \begin{array}{ccc} k & b+c & b^2+c^2 \\ 0 & a-b & a^2-b^2 \\ 0 & a-c & a^2-c^2 \end{array} \right|. \]

### Step 4: Factor Out Common Terms
We can factor \( a^2 - b^2 \) and \( a^2 - c^2 \):

\[ a^2 - b^2 = (a-b)(a+b), \]
\[ a^2 - c^2 = (a-c)(a+c). \]

Thus, we can rewrite the determinant as:

\[ D = k \cdot \left| \begin{array}{cc} a-b & (a-b)(a+b) \\ a-c & (a-c)(a+c) \end{array} \right|. \]

### Step 5: Calculate the 2x2 Determinant
The determinant of the 2x2 matrix is:

\[ \left| \begin{array}{cc} a-b & (a-b)(a+b) \\ a-c & (a-c)(a+c) \end{array} \right| = (a-b)(a-c) \left| \begin{array}{cc} 1 & a+b \\ 1 & a+c \end{array} \right|. \]

Calculating this gives:

\[ = (a-b)(a-c)((a+c) - (a+b)) = (a-b)(a-c)(c-b). \]

### Step 6: Final Expression for the Determinant
Thus, we have:

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

### Step 7: Set the Determinant Equal to the Right Side
We set this equal to the right side of the original equation:

\[ k \cdot (a-b)(a-c)(c-b) = (a-b)(b-c)(c-a). \]

### Step 8: Cancel Common Factors
Assuming \( a \neq b \), we can divide both sides by \( (a-b) \):

\[ k \cdot (a-c)(c-b) = (b-c)(c-a). \]

### Step 9: Solve for \( k \)
To find \( k \), we can rearrange:

\[ k = \frac{(b-c)(c-a)}{(a-c)(c-b)}. \]

### Step 10: Simplify and Determine \( k \)
Notice that the right-hand side simplifies to \( -1 \) when we consider the cyclic nature of the terms. Thus, we find:

\[ k = 1. \]

### Conclusion
The value of \( k \) is

\[ \boxed{1}. \]