If `x,y,zepsilonR` then the value of `|((2x^(x)+2^(-x))^(2),(2^(x)-2^(-x))^(2),1),((3x^(x)+3^(-x))^(2),(3^(x)-3^(-x))^(2),1),((4^(x)+4^(-x))^(2),(4^(x)-4^(-x))^(2),1)|` is

Explanation

To solve the determinant given in the question, we can follow these steps:

### Step 1: Write down the determinant
We start with the determinant:

\[ D = \begin{vmatrix} (2^x + 2^{-x})^2 & (2^x - 2^{-x})^2 & 1 \\ (3^x + 3^{-x})^2 & (3^x - 3^{-x})^2 & 1 \\ (4^x + 4^{-x})^2 & (4^x - 4^{-x})^2 & 1 \end{vmatrix} \]

### Step 2: Transform the first two columns
Notice that the first two columns have a specific structure. We can transform the first column (C1) by subtracting the second column (C2) from it:

\[ C1 \rightarrow C1 - C2 \]

This gives us:

\[ D = \begin{vmatrix} (2^x + 2^{-x})^2 - (2^x - 2^{-x})^2 & (2^x - 2^{-x})^2 & 1 \\ (3^x + 3^{-x})^2 - (3^x - 3^{-x})^2 & (3^x - 3^{-x})^2 & 1 \\ (4^x + 4^{-x})^2 - (4^x - 4^{-x})^2 & (4^x - 4^{-x})^2 & 1 \end{vmatrix} \]

### Step 3: Simplify the first column
Using the algebraic identity \( (a+b)^2 - (a-b)^2 = 4ab \), we can simplify each entry in the first column:

- For \( a = 2^x \) and \( b = 2^{-x} \):
\[ (2^x + 2^{-x})^2 - (2^x - 2^{-x})^2 = 4(2^x)(2^{-x}) = 4 \]

- For \( a = 3^x \) and \( b = 3^{-x} \):
\[ (3^x + 3^{-x})^2 - (3^x - 3^{-x})^2 = 4(3^x)(3^{-x}) = 4 \]

- For \( a = 4^x \) and \( b = 4^{-x} \):
\[ (4^x + 4^{-x})^2 - (4^x - 4^{-x})^2 = 4(4^x)(4^{-x}) = 4 \]

Thus, the determinant simplifies to:

\[ D = \begin{vmatrix} 4 & (2^x - 2^{-x})^2 & 1 \\ 4 & (3^x - 3^{-x})^2 & 1 \\ 4 & (4^x - 4^{-x})^2 & 1 \end{vmatrix} \]

### Step 4: Factor out the common term
We can factor out 4 from the first column:

\[ D = 4 \begin{vmatrix} 1 & (2^x - 2^{-x})^2 & 1 \\ 1 & (3^x - 3^{-x})^2 & 1 \\ 1 & (4^x - 4^{-x})^2 & 1 \end{vmatrix} \]

### Step 5: Identify identical rows
Now, observe that the first column consists of all 1's. If we subtract the first column from the second column, we will have:

\[ D = 4 \begin{vmatrix} 1 & (2^x - 2^{-x})^2 - 1 & 1 \\ 1 & (3^x - 3^{-x})^2 - 1 & 1 \\ 1 & (4^x - 4^{-x})^2 - 1 & 1 \end{vmatrix} \]

### Step 6: Conclude the determinant value
Since the first and third columns are now identical, the determinant is zero:

\[ D = 0 \]

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

\[ \boxed{0} \]