`|[-bc, b^2+bc, c^2 +bc],[a^2+ac,-ac,c^2+ac],[a^2+ab,b^2+ab,-ab]|=64`. then (ab+bc+ac) is

Answer

To solve the problem, we need to find the value of \( ab + bc + ac \) given that the determinant of the matrix

\[ \begin{bmatrix} -bc & b^2 + bc & c^2 + bc \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{bmatrix} \]

is equal to 64.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We denote the determinant of the matrix as \( D \):
\[ D = \begin{vmatrix} -bc & b^2 + bc & c^2 + bc \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{vmatrix} \]

2. **Apply Row Operations**:
We can simplify the determinant by multiplying the first row by \( a \), the second row by \( b \), and the third row by \( c \). This gives us:
\[ D = \frac{1}{abc} \begin{vmatrix} -abc & ab^2 + abc & ac^2 + abc \\ a^2b + abc & -abc & bc^2 + abc \\ a^2c + abc & b^2c + abc & -abc \end{vmatrix} \]

3. **Factor Out Common Terms**:
We can factor out \( a \) from the first column, \( b \) from the second column, and \( c \) from the third column:
\[ D = \frac{abc}{abc} \begin{vmatrix} -bc & ab + ac + bc & ac + ab \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{vmatrix} \]

4. **Simplify the Determinant**:
Now we can perform column operations to simplify the determinant further. We can replace the second column with the difference of the second and third columns:
\[ D = \begin{vmatrix} -bc & 0 & ac + ab \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{vmatrix} \]

5. **Calculate the Determinant**:
Now we can calculate the determinant using cofactor expansion or other methods. After simplification, we find that:
\[ D = (ab + ac + bc)^3 \]

6. **Set the Determinant Equal to 64**:
Given that \( D = 64 \), we have:
\[ (ab + ac + bc)^3 = 64 \]

7. **Solve for \( ab + ac + bc \)**:
Taking the cube root of both sides, we find:
\[ ab + ac + bc = 4 \]

### Final Answer:
Thus, the value of \( ab + ac + bc \) is \( \boxed{4} \).