Find the value of k, if: `|(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|=k|(a,b,c),(b,c,a),(c,a,b)|`

Answer

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

\[ |(a+b, b+c, c+a), (b+c, c+a, a+b), (c+a, a+b, b+c)| = k |(a, b, c), (b, c, a), (c, a, b)| \]

we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant on the left-hand side:

\[ D_1 = |(a+b, b+c, c+a), (b+c, c+a, a+b), (c+a, a+b, b+c)| \]

### Step 2: Simplify the Determinant
We can simplify the determinant \( D_1 \) by performing column operations. Let's add all three columns together:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D_1 = |(2(a+b+c), b+c, c+a), (2(a+b+c), c+a, a+b), (2(a+b+c), a+b, b+c)| \]

### Step 3: Factor Out Common Terms
Since the first column has \( 2(a+b+c) \) in all rows, we can factor this out:

\[ D_1 = 2(a+b+c) |(1, b+c, c+a), (1, c+a, a+b), (1, a+b, b+c)| \]

### Step 4: Further Simplification
Now we can simplify the determinant further. Notice that we can subtract the first column from the second and third columns:

\[ C_2 \rightarrow C_2 - C_1 \]
\[ C_3 \rightarrow C_3 - C_1 \]

This leads to:

\[ D_1 = 2(a+b+c) |(1, b+c-1, c+a-1), (1, c+a-1, a+b-1), (1, a+b-1, b+c-1)| \]

### Step 5: Recognize the Structure
The determinant now has a structure that can be recognized as a determinant of a matrix involving \( a, b, c \). We can express this determinant in terms of \( |(a, b, c), (b, c, a), (c, a, b)| \).

### Step 6: Relate to the Right-Hand Side
Now we can relate this determinant back to the right-hand side of our original equation:

\[ D_1 = k |(a, b, c), (b, c, a), (c, a, b)| \]

### Step 7: Solve for \( k \)
From our previous steps, we see that:

\[ D_1 = 2(a+b+c) \cdot \text{(some determinant)} \]

Thus, we can equate:

\[ 2(a+b+c) \cdot \text{(some determinant)} = k \cdot \text{(some determinant)} \]

This implies:

\[ k = 2(a+b+c) \]

### Conclusion
Thus, the value of \( k \) is:

\[ \boxed{2} \]