One of the roots of `|{:(x+a," "b," "c),(" "a,x+b," "c),(" "a," "b,x+c):}|=0` is :

Options

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

Explanation

To solve the problem, we need to find one of the roots of the determinant given by:

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

We need to set this determinant equal to zero and find the roots.

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

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

### Step 2: Apply column operations
Notice that the first column can be simplified. We can add the second and third columns to the first column:

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

This simplifies to:

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

### Step 3: Factor out common terms
We can factor out \(x + a + b + c\) from the first column:

\[ D = (x + a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & x + b & c \\ 0 & b & x + c \end{vmatrix} \]

### Step 4: Simplify the determinant
Now we can simplify the determinant further. We can perform row operations to make it easier to compute:

Subtract the first row from the second and third rows:

\[ D = (x + a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x + c - b \end{vmatrix} \]

### Step 5: Calculate the determinant
Now, we can calculate the determinant:

\[ D = (x + a + b + c) \cdot 1 \cdot x \cdot (x + c - b) \]

Thus, we have:

\[ D = (x + a + b + c) \cdot x \cdot (x + c - b) \]

### Step 6: Set the determinant to zero
To find the roots, we set the determinant equal to zero:

\[ (x + a + b + c) \cdot x \cdot (x + c - b) = 0 \]

### Step 7: Solve for roots
This gives us three possible roots:

1. \(x + a + b + c = 0 \Rightarrow x = - (a + b + c)\)
2. \(x = 0\)
3. \(x + c - b = 0 \Rightarrow x = b - c\)

### Conclusion
One of the roots of the determinant is:

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