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 given determinant, we will follow these steps:

Given 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 1: Simplify the determinant
We will simplify the first two columns of the determinant by subtracting the second column from the first column.

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

### Step 2: Apply the difference of squares
Using the identity \(a^2 - b^2 = (a + b)(a - b)\), we can simplify each entry in the first column:

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

### Step 3: Factor out the common term
Notice that we can factor out 4 from the first column:

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

### Step 4: Analyze the determinant
Now, we can see that the first column is proportional to the second column. Specifically, the first column can be expressed as a multiple of the second column. This means that the columns are linearly dependent.

### Step 5: Conclusion
If any two columns (or rows) of a determinant are identical or proportional, the value of the determinant is zero.

Thus, we conclude that:

\[ D = 0 \]

### Final Answer
The value of the determinant is \(0\).

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