If a,b,c be all +ive and p,q,r `in` R, then
`Delta=|((a^(p)+a^(-p))^(2),(a^(p)-a^(-p))^(2),1),((a^(q)+a^(-q))^(2),(a^(q)-a^(-q))^(2),1),((a^(r)+a^(-r))^(2),(a^(r)-a^(-r))^(2),1)|=`

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the given determinant \( \Delta \), we start with the expression:

\[ \Delta = \begin{vmatrix} (a^p + a^{-p})^2 & (a^p - a^{-p})^2 & 1 \\ (a^q + a^{-q})^2 & (a^q - a^{-q})^2 & 1 \\ (a^r + a^{-r})^2 & (a^r - a^{-r})^2 & 1 \end{vmatrix} \]

### Step 1: Expand the squares

We expand the squares in the determinant:

1. \((a^p + a^{-p})^2 = a^{2p} + 2 + a^{-2p}\)
2. \((a^p - a^{-p})^2 = a^{2p} - 2 + a^{-2p}\)
3. Similarly for \(q\) and \(r\).

Thus, we can rewrite the determinant as:

\[ \Delta = \begin{vmatrix} a^{2p} + 2 + a^{-2p} & a^{2p} - 2 + a^{-2p} & 1 \\ a^{2q} + 2 + a^{-2q} & a^{2q} - 2 + a^{-2q} & 1 \\ a^{2r} + 2 + a^{-2r} & a^{2r} - 2 + a^{-2r} & 1 \end{vmatrix} \]

### Step 2: Simplify the determinant

Now we can simplify the determinant by performing column operations. Let's perform the operation \(C_1 - C_2\):

\[ C_1 - C_2 = \begin{vmatrix} 4 & 4 & 1 \\ 4 & 4 & 1 \\ 4 & 4 & 1 \end{vmatrix} \]

This results in:

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

### Step 3: Evaluate the determinant

Notice that the first two rows of the determinant are now identical. Therefore, the determinant evaluates to zero:

\[ \Delta = 0 \]

### Conclusion

Thus, the value of the determinant \( \Delta \) is:

\[ \Delta = 0 \]