If a,b,c and d are the roots of the equation
`x^(4)+2x^(3)+4x^(2)+8x+16=0` the value of the determinant `|{:(1+a,1,1,1),(1,1+b,1,1),(1,1,1+c,1),(1,1,1,1+d):}|` is

Answer

To solve the problem, we need to find the value of the determinant

\[ D = \begin{vmatrix} 1 + a & 1 & 1 & 1 \\ 1 & 1 + b & 1 & 1 \\ 1 & 1 & 1 + c & 1 \\ 1 & 1 & 1 & 1 + d \end{vmatrix} \]

where \(a, b, c, d\) are the roots of the polynomial equation

\[ x^4 + 2x^3 + 4x^2 + 8x + 16 = 0. \]

### Step 1: Apply Row Operations
We can simplify the determinant using row operations. We perform the following operations:

- \(R_1 \leftarrow R_1 - R_4\)
- \(R_2 \leftarrow R_2 - R_4\)
- \(R_3 \leftarrow R_3 - R_4\)

This gives us:

\[ D = \begin{vmatrix} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d \end{vmatrix} \]

### Step 2: Calculate the Determinant
The determinant of a diagonal matrix is the product of its diagonal entries. Thus, we have:

\[ D = a \cdot b \cdot c \cdot d. \]

### Step 3: Use Vieta's Formulas
From the polynomial \(x^4 + 2x^3 + 4x^2 + 8x + 16 = 0\), we can use Vieta's formulas to find the relationships between the roots:

- The sum of the roots \(a + b + c + d = -2\).
- The sum of the products of the roots taken two at a time \(ab + ac + ad + bc + bd + cd = 4\).
- The sum of the products of the roots taken three at a time \(abc + abd + acd + bcd = -8\).
- The product of the roots \(abcd = 16\).

### Step 4: Substitute into the Determinant
Now we can substitute the value of the product of the roots into our determinant:

\[ D = abcd = 16. \]

### Final Answer
Thus, the value of the determinant is

\[ \boxed{16}. \]