`|((a^x+a^(-x))^2,(a^x-a^(-x))^(2),1),((b^x+b^(-x))^2,(b^x-b^(-x))^(2),1),((c^x+c^(-x))^2,(c^x-c^(-x))^(2),1)|` =

Options

A. 0
B. abc
C. 2abc
D. `a^2b^2c^2`

Answer

To solve the determinant

\[ D = \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (b^x + b^{-x})^2 & (b^x - b^{-x})^2 & 1 \\ (c^x + c^{-x})^2 & (c^x - c^{-x})^2 & 1 \end{vmatrix} \]

we will simplify the elements of the determinant step by step.

### Step 1: Simplify the first column
The first column consists of the terms \((a^x + a^{-x})^2\), \((b^x + b^{-x})^2\), and \((c^x + c^{-x})^2\).

Using the identity \((u + v)^2 = u^2 + v^2 + 2uv\), we can expand:

\[ (a^x + a^{-x})^2 = a^{2x} + a^{-2x} + 2 \]
\[ (b^x + b^{-x})^2 = b^{2x} + b^{-2x} + 2 \]
\[ (c^x + c^{-x})^2 = c^{2x} + c^{-2x} + 2 \]

Thus, the first column becomes:

\[ \begin{pmatrix} a^{2x} + a^{-2x} + 2 \\ b^{2x} + b^{-2x} + 2 \\ c^{2x} + c^{-2x} + 2 \end{pmatrix} \]

### Step 2: Simplify the second column
Now, we simplify the second column which consists of \((a^x - a^{-x})^2\), \((b^x - b^{-x})^2\), and \((c^x - c^{-x})^2\).

Using the identity \((u - v)^2 = u^2 + v^2 - 2uv\), we can expand:

\[ (a^x - a^{-x})^2 = a^{2x} + a^{-2x} - 2 \]
\[ (b^x - b^{-x})^2 = b^{2x} + b^{-2x} - 2 \]
\[ (c^x - c^{-x})^2 = c^{2x} + c^{-2x} - 2 \]

Thus, the second column becomes:

\[ \begin{pmatrix} a^{2x} + a^{-2x} - 2 \\ b^{2x} + b^{-2x} - 2 \\ c^{2x} + c^{-2x} - 2 \end{pmatrix} \]

### Step 3: Rewrite the determinant
Now we can rewrite the determinant \(D\):

\[ D = \begin{vmatrix} a^{2x} + a^{-2x} + 2 & a^{2x} + a^{-2x} - 2 & 1 \\ b^{2x} + b^{-2x} + 2 & b^{2x} + b^{-2x} - 2 & 1 \\ c^{2x} + c^{-2x} + 2 & c^{2x} + c^{-2x} - 2 & 1 \end{vmatrix} \]

### Step 4: Perform column operations
Now, we can perform the column operation \(C_1 \rightarrow C_1 - C_2\):

\[ D = \begin{vmatrix} 4 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix} \]

### Step 5: Evaluate the determinant
Notice that the first two columns are proportional (each row in the first column is a multiple of the corresponding row in the second column). Thus, the determinant evaluates to:

\[ D = 0 \]

### Final Answer
The value of the determinant is:

\[ \boxed{0} \]