If `|(2+x,x,x^(2)),(x,2+x,x^(2)),(x^(2),x,2+x)|=(1)/(6)(x-a)(x-b)(x-c)(x-d)` an identity in x where a, b, c, d are independent of x, then the value of `(13)/(25)abcd` is

Answer

To solve the problem, we need to evaluate the determinant and equate it to the given expression. Here is the step-by-step solution:

### Step 1: Write the Determinant
We need to evaluate the determinant of the matrix:
\[ D = \begin{vmatrix} 2+x & x & x^2 \\ x & 2+x & x^2 \\ x^2 & x & 2+x \end{vmatrix} \]

### Step 2: Expand the Determinant
Using the determinant formula for a 3x3 matrix, we can expand it as follows:
\[ D = (2+x) \begin{vmatrix} 2+x & x^2 \\ x & 2+x \end{vmatrix} - x \begin{vmatrix} x & x^2 \\ x^2 & 2+x \end{vmatrix} + x^2 \begin{vmatrix} x & 2+x \\ x^2 & x \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
1. For the first determinant:
\[ \begin{vmatrix} 2+x & x^2 \\ x & 2+x \end{vmatrix} = (2+x)(2+x) - x^2 \cdot x = (2+x)^2 - x^3 \]

2. For the second determinant:
\[ \begin{vmatrix} x & x^2 \\ x^2 & 2+x \end{vmatrix} = x(2+x) - x^2 \cdot x^2 = 2x + x^2 - x^4 \]

3. For the third determinant:
\[ \begin{vmatrix} x & 2+x \\ x^2 & x \end{vmatrix} = x \cdot x - (2+x)x^2 = x^2 - (2x^2 + x^3) = -2x^2 - x^3 + x^2 = -x^2 - x^3 \]

### Step 4: Substitute Back into the Determinant
Now substitute these back into the determinant:
\[ D = (2+x)((2+x)^2 - x^3) - x(2x + x^2 - x^4) + x^2(-x^2 - x^3) \]

### Step 5: Simplify the Expression
After expanding and simplifying the expression, we can equate it to the right-hand side of the equation:
\[ D = \frac{1}{6}(x-a)(x-b)(x-c)(x-d) \]

### Step 6: Evaluate at \(x = 0\)
To find the constants \(a, b, c, d\), we can evaluate the determinant at \(x = 0\):
\[ D(0) = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} = 2^3 = 8 \]
This gives:
\[ 8 = \frac{1}{6}(-a)(-b)(-c)(-d) = \frac{1}{6}abcd \]
Thus,
\[ abcd = 48 \]

### Step 7: Calculate \(\frac{13}{25} abcd\)
Now we can find the value of \(\frac{13}{25} abcd\):
\[ \frac{13}{25} \cdot 48 = \frac{624}{25} = 24.96 \]

### Final Answer
The value of \(\frac{13}{25} abcd\) is:
\[ \boxed{24.96} \]