Evaluate the following: ` |[b+c, a-b, a],[c+a, b-c, b],[a+b, c-a, c]| `

Options

A. `a^3+b^3 + c^3 -3abc`
B. `3abc-a^3-b^3-c^3`
C. `a^3+b^3 + c^3 -a^2b-b^2c-c^2a`
D. `(a+b+c)(a^2+b^2+c^2+ab+bc+ca)`

Answer

To evaluate the determinant

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

we will follow these steps:

### Step 1: Apply Column Operations
We will perform column operations to simplify the determinant. Specifically, we will add the third column to the first column and subtract the third column from the second column.

1. **Column 1**: \( C_1 \leftarrow C_1 + C_3 \)
2. **Column 2**: \( C_2 \leftarrow C_2 - C_3 \)

After performing these operations, the determinant becomes:

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

This simplifies to:

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

### Step 2: Factor Out Common Terms
Notice that the first column can be factored out since \( b+c+a \) is common in all rows:

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

### Step 3: Evaluate the 3x3 Determinant
Now we need to evaluate the determinant:

\[ D' = \begin{vmatrix} 1 & -b & a \\ 1 & -c & b \\ 1 & -a & c \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we can expand along the first row:

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

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} -c & b \\ -a & c \end{vmatrix} = (-c)(c) - (b)(-a) = -c^2 + ab\)

2. \(\begin{vmatrix} 1 & b \\ 1 & c \end{vmatrix} = (1)(c) - (1)(b) = c - b\)

3. \(\begin{vmatrix} 1 & -c \\ 1 & -a \end{vmatrix} = (1)(-a) - (1)(-c) = -a + c\)

Putting it all together:

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

### Step 4: Simplify the Expression
Now we simplify:

\[ D' = -c^2 + ab + bc - b^2 - a^2 + ac \]

### Step 5: Combine Terms
Rearranging gives:

\[ D' = ab + ac + bc - a^2 - b^2 - c^2 \]

### Step 6: Final Expression for D
Thus, we have:

\[ D = (b+c+a)(ab + ac + bc - a^2 - b^2 - c^2) \]

### Final Result
The final result is:

\[ D = (b+c+a)(ab + ac + bc - a^2 - b^2 - c^2) \]