What is `|{:(-a^(2),ab,ac),(ab,-b^(2),bc),(ac,bc,-c^(2)):}|` equal to ?

Options

A. 4 abc
B. `4a^(2)bc`
C. `4a^(2)b^(2)c^(2)`
D. `-4a^(2)b^(2)c^(2)`

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} \]

we can simplify the calculation by factoring out common terms from each row.

### Step 1: Factor out common terms from each row

From the first row, we can factor out \(-a\), from the second row, we can factor out \(-b\), and from the third row, we can factor out \(-c\):

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

This gives us:

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

### Step 2: Calculate the determinant of the simplified matrix

Now we need to calculate the determinant:

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

Using the determinant formula, we can expand this determinant:

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

### Step 3: Calculate the 2x2 determinants

1. For the first determinant:

\[ \begin{vmatrix} -b & c \\ c & -c \end{vmatrix} = (-b)(-c) - (c)(c) = bc - c^2 = c(b - c) \]

2. For the second determinant:

\[ \begin{vmatrix} b & c \\ c & -c \end{vmatrix} = (b)(-c) - (c)(c) = -bc - c^2 = -bc - c^2 \]

3. For the third determinant:

\[ \begin{vmatrix} b & -b \\ c & c \end{vmatrix} = (b)(c) - (-b)(c) = bc + bc = 2bc \]

### Step 4: Substitute back into the determinant equation

Now substituting these back into our determinant equation:

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

This simplifies to:

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

### Step 5: Combine like terms

Now we can combine the terms:

\[ D = acb - ac^2 + b^2c + bc^2 + 2bc^2 \]

This can be simplified further:

\[ D = acb - ac^2 + b^2c + 3bc^2 \]

### Final Result

Thus, the value of the determinant is:

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