`|[1,1,1] , [(2^x+2^(-x))^2, (3^x+3^(-x))^2, (5^x+5^(-x))^2] , [(2^x-2^(-x))^2, (3^x-3^(-x))^2, (5^x-5^(-x))^2]|=`

Options

A. 0
B. `30^x`
C. `30^(-x)`
D. None of these

Answer

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

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ (2^x + 2^{-x})^2 & (3^x + 3^{-x})^2 & (5^x + 5^{-x})^2 \\ (2^x - 2^{-x})^2 & (3^x - 3^{-x})^2 & (5^x - 5^{-x})^2 \end{vmatrix} \]

### Step 2: Expand the Squares
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\) and \((a - b)^2 = a^2 - 2ab + b^2\), we can expand the terms in the determinant:

- For \( (2^x + 2^{-x})^2 \):
\[ (2^x + 2^{-x})^2 = 2^{2x} + 2 + 2^{-2x} \]

- For \( (2^x - 2^{-x})^2 \):
\[ (2^x - 2^{-x})^2 = 2^{2x} - 2 + 2^{-2x} \]

- Similarly for \(3^x\) and \(5^x\):
\[ (3^x + 3^{-x})^2 = 3^{2x} + 2 + 3^{-2x} \]
\[ (3^x - 3^{-x})^2 = 3^{2x} - 2 + 3^{-2x} \]
\[ (5^x + 5^{-x})^2 = 5^{2x} + 2 + 5^{-2x} \]
\[ (5^x - 5^{-x})^2 = 5^{2x} - 2 + 5^{-2x} \]

### Step 3: Substitute Back into the Determinant
Now substitute these expansions back into the determinant:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2^{2x} + 2 + 2^{-2x} & 3^{2x} + 2 + 3^{-2x} & 5^{2x} + 2 + 5^{-2x} \\ 2^{2x} - 2 + 2^{-2x} & 3^{2x} - 2 + 3^{-2x} & 5^{2x} - 2 + 5^{-2x} \end{vmatrix} \]

### Step 4: Simplify the Determinant
Now we can simplify the determinant by performing row operations. We can subtract the second row from the third row:
\[ D = \begin{vmatrix} 1 & 1 & 1 \\ 2^{2x} + 2 + 2^{-2x} & 3^{2x} + 2 + 3^{-2x} & 5^{2x} + 2 + 5^{-2x} \\ 0 & 0 & 0 \end{vmatrix} \]

### Step 5: Evaluate the Determinant
Since one of the rows is now entirely zeros, the value of the determinant is:
\[ D = 0 \]

### Final Answer
Thus, the value of the determinant is:
\[ \boxed{0} \]