Using properties of determinants, prove that `|{:(,a,b,b+c),(,c,a,c+a),(,b,c,a+b):}|=(a+b+c) (a-c)^2`

Answer

To prove that

\[ \left| \begin{array}{ccc} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{array} \right| = (a+b+c)(a-c)^2, \]

we will use properties of determinants. Let's denote the determinant as \( D \).

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

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

### Step 2: Perform row operations
We will add the second and third rows to the first row. This operation does not change the value of the determinant.

\[ R_1 \to R_1 + R_2 + R_3 \]

This gives us:

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

### Step 3: Simplify the first row
Now, simplifying the first row:

\[ D = \left| \begin{array}{ccc} a + b + c & a + b + c & 2(a + b + c) \\ c & a & c + a \\ b & c & a + b \end{array} \right|. \]

### Step 4: Factor out the common term
We can factor out \( a + b + c \) from the first row:

\[ D = (a + b + c) \left| \begin{array}{ccc} 1 & 1 & 2 \\ c & a & c + a \\ b & c & a + b \end{array} \right|. \]

### Step 5: Perform column operations
Next, we will perform column operations to simplify further. We will subtract the first column from the second and subtract twice the first column from the third:

\[ C_2 \to C_2 - C_1, \quad C_3 \to C_3 - 2C_1. \]

This results in:

\[ D = (a + b + c) \left| \begin{array}{ccc} 1 & 0 & 0 \\ c & a - c & c + a - 2c \\ b & c - b & a + b - 2b \end{array} \right|. \]

### Step 6: Simplify the determinant
Now, simplifying the determinant further:

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

### Step 7: Calculate the determinant
The determinant simplifies to:

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

Calculating the 2x2 determinant:

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

### Step 8: Final expression
Thus, we have:

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

### Step 9: Rearranging
Rearranging gives us:

\[ D = (a + b + c)(a - c)^2. \]

### Conclusion
Thus, we have shown that:

\[ \left| \begin{array}{ccc} a & b & b+c \\ c & a & c+a \\ b & c & a+b \end{array} \right| = (a+b+c)(a-c)^2. \]

Hence proved.