If `a, b,c> 0` and `x,y,z in RR` 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)|` is equal to:

Options

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

Answer

To solve the determinant

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

we will follow these steps:

### Step 1: Simplify the elements of the determinant

We can denote:
- \( A = a^x \)
- \( B = b^y \)
- \( C = c^z \)

Thus, we can rewrite the determinant as:

\[ D = \begin{vmatrix} (A + A^{-1})^2 & (A - A^{-1})^2 & 1 \\ (B + B^{-1})^2 & (B - B^{-1})^2 & 1 \\ (C + C^{-1})^2 & (C - C^{-1})^2 & 1 \end{vmatrix} \]

### Step 2: Use properties of determinants

We can use the property of determinants that allows us to perform column operations. Specifically, we can subtract the second column from the first column:

\[ D = \begin{vmatrix} (A + A^{-1})^2 - (A - A^{-1})^2 & (A - A^{-1})^2 & 1 \\ (B + B^{-1})^2 - (B - B^{-1})^2 & (B - B^{-1})^2 & 1 \\ (C + C^{-1})^2 - (C - C^{-1})^2 & (C - C^{-1})^2 & 1 \end{vmatrix} \]

### Step 3: Simplify the first column

Now we simplify the first column:

\[ (A + A^{-1})^2 - (A - A^{-1})^2 = (A^2 + 2 + A^{-2}) - (A^2 - 2 + A^{-2}) = 4 \]

Thus, the first column becomes \( [4, 4, 4] \):

\[ D = \begin{vmatrix} 4 & (A - A^{-1})^2 & 1 \\ 4 & (B - B^{-1})^2 & 1 \\ 4 & (C - C^{-1})^2 & 1 \end{vmatrix} \]

### Step 4: Factor out the common term

Since the first column is now constant (all entries are 4), we can factor out 4:

\[ D = 4 \begin{vmatrix} 1 & (A - A^{-1})^2 & 1 \\ 1 & (B - B^{-1})^2 & 1 \\ 1 & (C - C^{-1})^2 & 1 \end{vmatrix} \]

### Step 5: Recognize the determinant structure

Notice that the first and third columns are identical. Therefore, the determinant evaluates to zero:

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

### Conclusion

Thus, the value of the determinant is:

\[ \boxed{0} \]