Use properties of determinants to solve for x: `|{:(,x+a,b,c),(,c,x+b,a),(,a,b,x+c):}|=0 and x=0`

Answer

To solve the determinant equation

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

we will use properties of determinants step by step.

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

The new first column becomes:
- First row: \( (x + a) + b + c = x + a + b + c \)
- Second row: \( c + (x + b) + a = x + a + b + c \)
- Third row: \( a + b + (x + c) = x + a + b + c \)

Thus, the determinant transforms to:

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

### Step 2: Subtract Rows
Next, we will perform row operations. We will subtract the second row from the first row and the third row from the second row.

1. \( R_1 = R_1 - R_2 \)
2. \( R_2 = R_2 - R_3 \)

After these operations, we get:

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

This simplifies to:

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

### Step 3: Expand the Determinant
Now we can expand the determinant along the first column. The first column has zeros, so we can ignore it:

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

Calculating the 2x2 determinant:

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

This simplifies to:

\[ = (x + a + b + c) \left( -ax + x^2 + xc - cx + ac - a^2 \right) \]

### Step 4: Set the Determinant to Zero
Now we set the entire expression to zero:

\[ (x + a + b + c) \left( -ax + x^2 + xc - cx + ac - a^2 \right) = 0 \]

This gives us two cases:
1. \( x + a + b + c = 0 \)
2. The quadratic equation \( -ax + x^2 + xc - cx + ac - a^2 = 0 \)

### Step 5: Solve for x
From the first case, we have:

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

From the second case, we can factor or use the quadratic formula to find the roots. However, we already know that one solution is \( x = 0 \).

### Final Solutions
Thus, the final solutions for \( x \) are:

\[ x = 0 \quad \text{or} \quad x = - (a + b + c) \]