If `|(alpha^(2n),alpha^(2n+2),alpha^(2n+4)),(beta^(2n),beta^(2n+2),beta^(2n+4)),(gamma^(2n),gamma^(2n+2),gamma^(2n+4))|=((1)/(beta^(2))-(1)/(alpha^(2)))((1)/(gamma^(2))-(1)/(beta^(2)))((1)/(alpha^(2))-(1)/(gamma^(2)))`
{where `alpha^(2), beta^(2) and gamma^(2)` are al distinct}, then the value of n is equal to

Options

A. 4
B. `-4`
C. 3
D. `-2`

Answer

To solve the problem, we need to calculate the determinant on the left-hand side and equate it to the expression on the right-hand side. Here’s the step-by-step solution:

### Step 1: Factor out common terms from the determinant

We have:
\[ \text{Det} = \begin{vmatrix} \alpha^{2n} & \alpha^{2n+2} & \alpha^{2n+4} \\ \beta^{2n} & \beta^{2n+2} & \beta^{2n+4} \\ \gamma^{2n} & \gamma^{2n+2} & \gamma^{2n+4} \end{vmatrix} \]

We can factor out \(\alpha^{2n}\), \(\beta^{2n}\), and \(\gamma^{2n}\) from each row respectively:
\[ = \alpha^{2n} \beta^{2n} \gamma^{2n} \begin{vmatrix} 1 & \alpha^2 & \alpha^4 \\ 1 & \beta^2 & \beta^4 \\ 1 & \gamma^2 & \gamma^4 \end{vmatrix} \]

### Step 2: Evaluate the determinant

The determinant can be evaluated using the formula for a 3x3 determinant:
\[ \begin{vmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{vmatrix} = (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) \]

In our case, \(x_1 = \alpha^2\), \(x_2 = \beta^2\), \(x_3 = \gamma^2\). Thus,
\[ \begin{vmatrix} 1 & \alpha^2 & \alpha^4 \\ 1 & \beta^2 & \beta^4 \\ 1 & \gamma^2 & \gamma^4 \end{vmatrix} = (\alpha^2 - \beta^2)(\beta^2 - \gamma^2)(\gamma^2 - \alpha^2) \]

### Step 3: Combine results

Now substituting back, we have:
\[ \text{Det} = \alpha^{2n} \beta^{2n} \gamma^{2n} \cdot (\alpha^2 - \beta^2)(\beta^2 - \gamma^2)(\gamma^2 - \alpha^2) \]

### Step 4: Evaluate the right-hand side

The right-hand side is given by:
\[ \left(\frac{1}{\beta^2} - \frac{1}{\alpha^2}\right)\left(\frac{1}{\gamma^2} - \frac{1}{\beta^2}\right)\left(\frac{1}{\alpha^2} - \frac{1}{\gamma^2}\right) \]

Taking the LCM for each term:
\[ = \frac{\alpha^2 - \beta^2}{\alpha^2 \beta^2} \cdot \frac{\beta^2 - \gamma^2}{\beta^2 \gamma^2} \cdot \frac{\gamma^2 - \alpha^2}{\gamma^2 \alpha^2} \]

### Step 5: Combine and equate

Now we can equate both sides:
\[ \alpha^{2n} \beta^{2n} \gamma^{2n} \cdot (\alpha^2 - \beta^2)(\beta^2 - \gamma^2)(\gamma^2 - \alpha^2) = \frac{(\alpha^2 - \beta^2)(\beta^2 - \gamma^2)(\gamma^2 - \alpha^2)}{\alpha^2 \beta^2 \gamma^2} \]

### Step 6: Simplify

Canceling \((\alpha^2 - \beta^2)(\beta^2 - \gamma^2)(\gamma^2 - \alpha^2)\) from both sides (since they are non-zero):
\[ \alpha^{2n} \beta^{2n} \gamma^{2n} = \frac{1}{\alpha^2 \beta^2 \gamma^2} \]

### Step 7: Solve for \(n\)

This implies:
\[ \alpha^{2n} \beta^{2n} \gamma^{2n} = \alpha^{-2} \beta^{-2} \gamma^{-2} \]

Taking logarithm or comparing powers, we get:
\[ 2n = -2 \implies n = -1 \]

### Final Answer

Thus, the value of \(n\) is:
\[ \boxed{-1} \]