If `alpha, beta, gamma and delta` are the roots of the equation `x ^(4) -bx -3 =0,` then an equation whose roots are `(alpha +beta+gamma)/(delta^(2)), (alpha +beta+delta)/(gamma^(2)), (alpha +delta+gamma)/(beta^(2)), and (delta +beta+gamma)/(alpha^(2)), ` is:

Options

A. `3x ^(4) +bx+1=0`
B. `3x ^(4) -bx +1=0`
C. `3x ^(4) +bx ^(3) -1=0`
D. `3x ^(4) -bx ^(3) -1=0`

Answer

To solve the problem, we need to find an equation whose roots are given in terms of the roots of the original polynomial \( x^4 - bx - 3 = 0 \). Let's denote the roots of the original equation as \( \alpha, \beta, \gamma, \delta \).

### Step 1: Find the sum of the roots
From Vieta's formulas, we know that for a polynomial of the form \( x^4 + 0x^3 + 0x^2 - bx - 3 = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma + \delta = 0 \).

### Step 2: Express \( \alpha + \beta + \gamma \)
Using the result from Step 1:
\[ \alpha + \beta + \gamma = -\delta \]

### Step 3: Calculate the new roots
The new roots are given by:
1. \( \frac{\alpha + \beta + \gamma}{\delta^2} = \frac{-\delta}{\delta^2} = -\frac{1}{\delta} \)
2. \( \frac{\alpha + \beta + \delta}{\gamma^2} = \frac{-\gamma}{\gamma^2} = -\frac{1}{\gamma} \)
3. \( \frac{\alpha + \delta + \gamma}{\beta^2} = \frac{-\beta}{\beta^2} = -\frac{1}{\beta} \)
4. \( \frac{\delta + \beta + \gamma}{\alpha^2} = \frac{-\alpha}{\alpha^2} = -\frac{1}{\alpha} \)

Thus, the new roots can be expressed as:
\[ -\frac{1}{\delta}, -\frac{1}{\gamma}, -\frac{1}{\beta}, -\frac{1}{\alpha} \]

### Step 4: Form the new polynomial
The roots of the new polynomial can be represented as:
\[ x = -\frac{1}{\alpha} \Rightarrow \alpha = -\frac{1}{x} \]
Substituting \( x \) with \( -\frac{1}{x} \) in the original polynomial \( x^4 - bx - 3 = 0 \):
\[ \left(-\frac{1}{x}\right)^4 - b\left(-\frac{1}{x}\right) - 3 = 0 \]
This simplifies to:
\[ \frac{1}{x^4} + \frac{b}{x} - 3 = 0 \]
Multiplying through by \( x^4 \) gives:
\[ 1 + bx^3 - 3x^4 = 0 \]
Rearranging gives:
\[ 3x^4 - bx^3 - 1 = 0 \]

### Final Result
The equation whose roots are \( -\frac{1}{\delta}, -\frac{1}{\gamma}, -\frac{1}{\beta}, -\frac{1}{\alpha} \) is:
\[ 3x^4 - bx^3 - 1 = 0 \]