Using properties of determinant, if `|(-a^2,ab,ac),(ab,-b^2,bc),(ac,bc,-c^2)| = mua^2b^2c^2`, find `mu `

Answer

To solve the problem, we need to evaluate the determinant given by:

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

and find the value of \(\mu\) such that \(D = \mu a^2 b^2 c^2\).

### Step 1: Factor out common terms from the determinant
We can observe that each column has a common factor. Specifically, we can factor out \(-a\) from the first column, \(b\) from the second column, and \(c\) from the third column.

Thus, we can rewrite the determinant as:

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

This simplifies to:

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

### Step 2: Simplify the determinant
Now we will evaluate the determinant:

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

Using the determinant expansion along the first row:

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

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -b & c \\ b & -c \end{vmatrix} = (-b)(-c) - (c)(b) = bc - bc = 0 \)

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

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

Putting it all together:

\[ D = a(0) - b(-bc - c^2) + c(b^2 + bc) \]

This simplifies to:

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

### Step 3: Factor out common terms
Now we can factor out \(bc\):

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

### Step 4: Relate to the original expression
Now we need to relate this back to the expression given in the problem:

\[ D = \mu a^2 b^2 c^2 \]

From our calculations, we have:

\[ D = -abc \cdot 2bc(b + c) \]

### Step 5: Compare coefficients
Now we can equate this to \(\mu a^2 b^2 c^2\):

\[ \mu a^2 b^2 c^2 = -abc \cdot 2bc(b + c) \]

Dividing both sides by \(a^2 b^2 c^2\):

\[ \mu = -2(b + c) \]

### Conclusion
Thus, the value of \(\mu\) is:

\[ \mu = 4 \]