Without expanding the determinant, show that `(a+b+c)` is a factor of the determinant `|a b c b c a c a b|` .

Explanation

To show that \( (a + b + c) \) is a factor of the determinant

\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

we will use column operations without expanding the determinant.

### Step 1: Perform Column Operation
We will perform the column operation \( C_1 \to C_1 + C_2 + C_3 \). This means we will add the second and third columns to the first column.

After performing this operation, the first column becomes:

\[ C_1 = a + b + c \]

The second and third columns remain unchanged:

\[ D = \begin{vmatrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{vmatrix} \]

### Step 2: Factor Out Common Column
Now we can see that the first column has the common factor \( (a + b + c) \). We can factor this out of the determinant:

\[ D = (a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

### Step 3: Conclusion
Since we have factored out \( (a + b + c) \) from the determinant, we conclude that \( (a + b + c) \) is indeed a factor of the determinant \( D \).