Solve the equation `|{:(x+a,x+b,x+c),(x+b,x+c,x+a),(x+c,x+a,x+b):}|=0`

Answer

To solve the equation given by the determinant

\[ \left| \begin{array}{ccc} x+a & x+b & x+c \\ x+b & x+c & x+a \\ x+c & x+a & x+b \end{array} \right| = 0, \]

we will follow these steps:

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

\[ D = \left| \begin{array}{ccc} x+a & x+b & x+c \\ x+b & x+c & x+a \\ x+c & x+a & x+b \end{array} \right| \]

### Step 2: Apply Row Transformation
We can simplify the determinant using the property of determinants that allows us to add rows. We can add all three rows together to the first row:

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives:

\[ D = \left| \begin{array}{ccc} 3x + a + b + c & x+b & x+c \\ x+b & x+c & x+a \\ x+c & x+a & x+b \end{array} \right| \]

### Step 3: Factor Out Common Terms
Now, we can factor out \(3x + a + b + c\) from the first row:

\[ D = (3x + a + b + c) \left| \begin{array}{ccc} 1 & 1 & 1 \\ x+b & x+c & x+a \\ x+c & x+a & x+b \end{array} \right| \]

### Step 4: Apply Column Transformation
Next, we can perform column operations to simplify the determinant further. We can subtract the second column from the first and the third from the second:

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

This results in:

\[ D = (3x + a + b + c) \left| \begin{array}{ccc} 0 & 1 & 1 \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{array} \right| \]

### Step 5: Calculate the Remaining Determinant
Now we can calculate the determinant of the 3x3 matrix. The first row has a zero, so we can expand along that row:

\[ D = (3x + a + b + c) \cdot 0 + 1 \cdot \left| \begin{array}{cc} b - c & a - b \\ c - a & b - c \end{array} \right| - 1 \cdot \left| \begin{array}{cc} b - c & c - a \\ c - a & a - b \end{array} \right| \]

### Step 6: Set the Determinant to Zero
Now we set the determinant \(D\) to zero:

\[ (3x + a + b + c) \cdot \text{determinant} = 0 \]

This gives us two cases to solve:

1. \(3x + a + b + c = 0\)
2. The determinant of the 2x2 matrix must also equal zero.

### Step 7: Solve for \(x\)
From the first equation:

\[ 3x + a + b + c = 0 \implies 3x = - (a + b + c) \implies x = -\frac{a + b + c}{3} \]

### Step 8: Conclude
Thus, the solution for \(x\) is:

\[ x = -\frac{a + b + c}{3} \]