In a square matrix A of order 3 the elements `a_(ij)` 's are the
sum of the roots of the equation `x^(2) - (a+b) x + ab =0,`
`a_(I,i+1)`'s are the product of the roots, `a_(I,i-1)` 's are all unity
and the rest of the elements are all zero. The value of the det (A) is equal to

Explanation

To solve the given problem, we need to construct the matrix \( A \) based on the provided conditions and then calculate its determinant. Let's go through the steps systematically.

### Step 1: Understand the roots of the quadratic equation
The given quadratic equation is:
\[ x^2 - (a+b)x + ab = 0 \]
The sum of the roots \( S \) of this equation is given by:
\[ S = a + b \]
The product of the roots \( P \) is given by:
\[ P = ab \]

### Step 2: Construct the matrix \( A \)
The matrix \( A \) of order 3 can be represented as follows based on the conditions provided:
- The elements \( a_{ij} \) are the sum of the roots, which is \( a + b \).
- The elements \( a_{i,i+1} \) (the element to the right of the diagonal) are the product of the roots, which is \( ab \).
- The elements \( a_{i,i-1} \) (the element to the left of the diagonal) are all unity (1).
- All other elements are zero.

Thus, the matrix \( A \) can be structured as:
\[ A = \begin{bmatrix} a+b & ab & 0 \\ 1 & a+b & ab \\ 0 & 1 & a+b \end{bmatrix} \]

### Step 3: Calculate the determinant of matrix \( A \)
To find the determinant of \( A \), we can use the formula for the determinant of a 3x3 matrix:
\[ \text{det}(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]
Substituting the elements from matrix \( A \):
\[ \text{det}(A) = (a+b) \left( (a+b)(a+b) - ab \cdot 1 \right) - ab \left( 1 \cdot (a+b) - ab \cdot 0 \right) + 0 \]
This simplifies to:
\[ \text{det}(A) = (a+b) \left( (a+b)^2 - ab \right) - ab(a+b) \]
Expanding \( (a+b)^2 - ab \):
\[ (a+b)^2 - ab = a^2 + 2ab + b^2 - ab = a^2 + ab + b^2 \]
Thus, we have:
\[ \text{det}(A) = (a+b)(a^2 + ab + b^2) - ab(a+b) \]
Factoring out \( (a+b) \):
\[ \text{det}(A) = (a+b) \left( a^2 + ab + b^2 - ab \right) = (a+b)(a^2 + b^2) \]
Since \( a^2 + b^2 = (a+b)^2 - 2ab \), we can rewrite it as:
\[ \text{det}(A) = (a+b)((a+b)^2 - 2ab) \]
This leads us to the final determinant:
\[ \text{det}(A) = (a+b)^3 \]

### Final Answer
\[ \text{det}(A) = (a+b)^3 \]