Let A be a square matrix of order 3 such that `A=A^(T)=[(10,4,6),(a_(21)+a_(12),6,a_(23)+a_(32)),(a_(31)+a_(13),8,4)]`,where `a_(12), a_(23), a_(31)` are positive roots of the equation `x^(3)-6x^(2)+px-8=0, AA p in R`, then the absolute vlaue of `|A|` is equal to

Answer

To solve the problem, we need to find the absolute value of the determinant of the symmetric matrix \( A \) given by:

\[ A = \begin{pmatrix} 10 & 4 & 6 \\ a_{21} + a_{12} & 6 & a_{23} + a_{32} \\ a_{31} + a_{13} & 8 & 4 \end{pmatrix} \]

where \( a_{12}, a_{23}, a_{31} \) are the positive roots of the polynomial equation \( x^3 - 6x^2 + px - 8 = 0 \).

### Step 1: Identify the roots and their properties
From Vieta's formulas, we know:
- The sum of the roots \( a_{12} + a_{23} + a_{31} = 6 \)
- The product of the roots \( a_{12} a_{23} a_{31} = 8 \)

### Step 2: Assume values for the roots
Since \( a_{12}, a_{23}, a_{31} \) are positive roots and their sum is 6, we can assume:
\[ a_{12} = a_{23} = a_{31} = 2 \]
This satisfies both the sum and the product conditions:
- \( 2 + 2 + 2 = 6 \)
- \( 2 \cdot 2 \cdot 2 = 8 \)

### Step 3: Substitute the values into the matrix
Now substituting \( a_{12} = 2, a_{23} = 2, a_{31} = 2 \) into the matrix \( A \):
\[ A = \begin{pmatrix} 10 & 4 & 6 \\ 2 + 2 & 6 & 2 + 2 \\ 2 + 2 & 8 & 4 \end{pmatrix} = \begin{pmatrix} 10 & 4 & 6 \\ 4 & 6 & 4 \\ 4 & 8 & 4 \end{pmatrix} \]

### Step 4: Calculate the determinant of matrix \( A \)
To calculate the determinant of \( A \):
\[ |A| = \begin{vmatrix} 10 & 4 & 6 \\ 4 & 6 & 4 \\ 4 & 8 & 4 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \).

Substituting the values:
\[ |A| = 10(6 \cdot 4 - 4 \cdot 8) - 4(4 \cdot 4 - 4 \cdot 8) + 6(4 \cdot 8 - 6 \cdot 4) \]
Calculating each term:
1. \( 6 \cdot 4 - 4 \cdot 8 = 24 - 32 = -8 \)
2. \( 4 \cdot 4 - 4 \cdot 8 = 16 - 32 = -16 \)
3. \( 4 \cdot 8 - 6 \cdot 4 = 32 - 24 = 8 \)

Now substituting back:
\[ |A| = 10(-8) - 4(-16) + 6(8) \]
Calculating:
\[ |A| = -80 + 64 + 48 = 32 \]

### Step 5: Absolute value of the determinant
The absolute value of the determinant is:
\[ |A| = 32 \]

### Final Answer
Thus, the absolute value of \( |A| \) is \( \boxed{32} \).