What is the value of the determinant
`|(a-b,b+c,a),(b-c,c+a,b),(c-a,a+b,c)|` ?

Options

A. `a^(3) + b^(3) + c^(3)`
B. 3bc
C. `a^(3) + b^(3) + c^(3) - 3abc`
D. 0

Answer

To find the value of the determinant

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

we will perform a series of column operations to simplify the determinant.

### Step 1: Perform Column Operation
We will start by performing the column operation \( C_1 \to C_1 - C_3 \).

This gives us:

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

Simplifying the first column:

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

### Step 2: Perform Another Column Operation
Next, we will perform the column operation \( C_2 \to C_2 + C_1 \).

This gives us:

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

Simplifying the second column:

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

### Step 3: Expand the Determinant
Now we will expand the determinant along the first row:

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

Calculating the 2x2 determinants:

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

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

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

Substituting these back into the expression for \(D\):

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

### Step 4: Simplify the Expression
Expanding this gives:

\[ D = -abc + b^3 - c^3 + abc + a^3 - abc \]

Combining like terms results in:

\[ D = a^3 + b^3 + c^3 - 3abc \]

### Final Result
Thus, the value of the determinant is:

\[ \boxed{a^3 + b^3 + c^3 - 3abc} \]

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