The root of the equation `|(x,alpha,1),(beta,x,1),(beta,gamma,1)| = 0` are independent of :

Options

A. (a) `alpha`
B. (b) `beta`
C. (c) `gamma`
D. (d) `alpha, beta and gamma`

Answer

To solve the determinant equation \(|(x, \alpha, 1), (\beta, x, 1), (\beta, \gamma, 1)| = 0\) and determine which variable the roots are independent of, we can follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ D = \begin{vmatrix} x & \alpha & 1 \\ \beta & x & 1 \\ \beta & \gamma & 1 \end{vmatrix} \]

### Step 2: Expand the Determinant
Using the first row to expand the determinant:
\[ D = x \begin{vmatrix} x & 1 \\ \gamma & 1 \end{vmatrix} - \alpha \begin{vmatrix} \beta & 1 \\ \beta & 1 \end{vmatrix} + 1 \begin{vmatrix} \beta & x \\ \beta & \gamma \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinants
Now we calculate each of the 2x2 determinants:
1. For the first determinant:
\[ \begin{vmatrix} x & 1 \\ \gamma & 1 \end{vmatrix} = x \cdot 1 - \gamma \cdot 1 = x - \gamma \]

2. For the second determinant:
\[ \begin{vmatrix} \beta & 1 \\ \beta & 1 \end{vmatrix} = \beta \cdot 1 - \beta \cdot 1 = 0 \]

3. For the third determinant:
\[ \begin{vmatrix} \beta & x \\ \beta & \gamma \end{vmatrix} = \beta \cdot \gamma - \beta \cdot x = \beta(\gamma - x) \]

### Step 4: Substitute Back into the Determinant
Substituting these results back into the determinant:
\[ D = x(x - \gamma) - \alpha \cdot 0 + \beta(\gamma - x) \]
This simplifies to:
\[ D = x^2 - x\gamma + \beta\gamma - \beta x \]
Rearranging gives:
\[ D = x^2 - (\gamma + \beta)x + \beta\gamma \]

### Step 5: Set the Determinant to Zero
For the determinant to be zero:
\[ x^2 - (\gamma + \beta)x + \beta\gamma = 0 \]

### Step 6: Analyze the Roots
The roots of the quadratic equation are given by:
\[ x = \frac{(\gamma + \beta) \pm \sqrt{(\gamma + \beta)^2 - 4\beta\gamma}}{2} \]
The roots depend on the coefficients \(\gamma\) and \(\beta\), but do not depend on \(\alpha\).

### Conclusion
Thus, the roots of the equation are independent of \(\alpha\).