If `a, b,c gt 0& x,y,z in R" then the determinant"[{:(,(a^(x)+a^(-x))^(2),(a^(x)-a^(-x))^(2),1),(,(b^(y)+b^(-y))^(2),(b^(y)-b^(-y))^(2),1),(,(c^(z)+c^(-z))^(2),(c^(z)-c^(-z))^(2),1):}]=`

Options

A. `a^(x)b^(y)c^(z)`
B. `a^(-x)b^(-y)c^(-z)`
C. `a^(2x)b^(2y)c^(2z)`
D. zero

Answer

To solve the determinant given in the question, we will follow a systematic approach. The determinant is given by:

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

### Step 1: Expand the terms in the determinant

First, we will expand the squares in the determinant.

1. For the first row:
\[ (a^x + a^{-x})^2 = a^{2x} + 2 + a^{-2x} \]
\[ (a^x - a^{-x})^2 = a^{2x} - 2 + a^{-2x} \]

2. For the second row:
\[ (b^y + b^{-y})^2 = b^{2y} + 2 + b^{-2y} \]
\[ (b^y - b^{-y})^2 = b^{2y} - 2 + b^{-2y} \]

3. For the third row:
\[ (c^z + c^{-z})^2 = c^{2z} + 2 + c^{-2z} \]
\[ (c^z - c^{-z})^2 = c^{2z} - 2 + c^{-2z} \]

Thus, the determinant becomes:

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

### Step 2: Simplify the determinant

Next, we will simplify the determinant by subtracting the second column from the first column:

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

This simplifies to:

\[ D = \begin{vmatrix} 4 & a^{2x} - 2 + a^{-2x} & 1 \\ 4 & b^{2y} - 2 + b^{-2y} & 1 \\ 4 & c^{2z} - 2 + c^{-2z} & 1 \end{vmatrix} \]

### Step 3: Factor out the common term

Now, we can factor out the common term (4) from the first column:

\[ D = 4 \begin{vmatrix} 1 & a^{2x} - 2 + a^{-2x} & 1 \\ 1 & b^{2y} - 2 + b^{-2y} & 1 \\ 1 & c^{2z} - 2 + c^{-2z} & 1 \end{vmatrix} \]

### Step 4: Evaluate the determinant

Notice that the first column now has identical entries (all 1s). Therefore, the determinant evaluates to zero:

\[ D = 4 \cdot 0 = 0 \]

### Final Answer

Thus, the value of the determinant is:

\[ \boxed{0} \]