If ` alpha , beta , gamma ` are the roots of ` x^3 -3x +1=0` then the equation whose roots are
`alpha - (1)/( beta gamma) , beta - (1)/( gamma alpha ) , gamma - (1)/( alpha beta )` is

Options

A. `x^3 - 3x +8=0`
B. ` x^3 -6x +8=0`
C. `x^3 -9x +8=0`
D. `x^3 - 12 x + 8=0`

Answer

To find the equation whose roots are \( \alpha - \frac{1}{\beta \gamma}, \beta - \frac{1}{\gamma \alpha}, \gamma - \frac{1}{\alpha \beta} \), given that \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 - 3x + 1 = 0 \), we will follow these steps:

### Step 1: Identify the coefficients from the original polynomial
The polynomial is given as:
\[ x^3 - 3x + 1 = 0 \]
From this, we can identify:
- \( a = 1 \)
- \( b = 0 \)
- \( c = -3 \)
- \( d = 1 \)

### Step 2: Use Vieta's formulas
Using Vieta's formulas, we can find:
1. \( \alpha + \beta + \gamma = -\frac{b}{a} = 0 \)
2. \( \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} = -3 \)
3. \( \alpha \beta \gamma = -\frac{d}{a} = -1 \)

### Step 3: Simplify the new roots
We need to simplify each of the new roots:
1. \( \alpha - \frac{1}{\beta \gamma} = \alpha - \frac{1}{-1} = \alpha + 1 \)
2. \( \beta - \frac{1}{\gamma \alpha} = \beta - \frac{1}{-1} = \beta + 1 \)
3. \( \gamma - \frac{1}{\alpha \beta} = \gamma - \frac{1}{-1} = \gamma + 1 \)

Thus, the new roots are \( \alpha + 1, \beta + 1, \gamma + 1 \).

### Step 4: Find the new polynomial
Let \( y = x - 1 \) (i.e., \( x = y + 1 \)). Then the roots \( \alpha + 1, \beta + 1, \gamma + 1 \) correspond to \( y = \alpha, \beta, \gamma \).

Substituting \( x = y + 1 \) into the original polynomial:
\[ (y + 1)^3 - 3(y + 1) + 1 = 0 \]
Expanding this:
\[ y^3 + 3y^2 + 3y + 1 - 3y - 3 + 1 = 0 \]
Simplifying:
\[ y^3 + 3y^2 - 1 = 0 \]

### Step 5: Write the final polynomial
Thus, the polynomial whose roots are \( \alpha + 1, \beta + 1, \gamma + 1 \) is:
\[ y^3 + 3y^2 - 1 = 0 \]
Substituting back \( y = x - 1 \):
\[ x^3 + 3x^2 - 1 = 0 \]

### Final Answer
The required equation is:
\[ x^3 + 3x^2 - 1 = 0 \]