If ` alpha , beta , gamma ` are the roots of ` x^3 +x^2 + 2x +3=0` where `a+b+c = -2` , `ab+bc+ca=3`, `abc=1` then the equation whose roots ` beta + gamma , gamma + alpha , alpha + beta ` is

Options

A. `x^3 +2x^2 -3x +1=0`
B. `x^3 +2x^2 +3x +1=0`
C. `x^3 +2x^2 +3x +1=0`
D. `x^3 +2x^2 +3x -1=0`

Answer

To find the equation whose roots are \( \beta + \gamma, \gamma + \alpha, \alpha + \beta \) given that \( \alpha, \beta, \gamma \) are the roots of the polynomial \( x^3 + x^2 + 2x + 3 = 0 \), we can follow these steps:

### Step 1: Identify the sums of the roots
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -1 \).
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = 2 \).
- The product of the roots \( \alpha\beta\gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -3 \).

### Step 2: Calculate the new roots
Let:
- \( a = \beta + \gamma \)
- \( b = \gamma + \alpha \)
- \( c = \alpha + \beta \)

Now, we can express the sums and products of \( a, b, c \):
- \( a + b + c = (\beta + \gamma) + (\gamma + \alpha) + (\alpha + \beta) = 2(\alpha + \beta + \gamma) = 2(-1) = -2 \)
- \( ab + bc + ca = (\beta + \gamma)(\gamma + \alpha) + (\gamma + \alpha)(\alpha + \beta) + (\alpha + \beta)(\beta + \gamma) \)

Calculating \( ab + bc + ca \):
\[ ab = (\beta + \gamma)(\gamma + \alpha) = \beta\gamma + \beta\alpha + \gamma^2 + \gamma\alpha \]
\[ bc = (\gamma + \alpha)(\alpha + \beta) = \gamma\alpha + \gamma\beta + \alpha^2 + \alpha\beta \]
\[ ca = (\alpha + \beta)(\beta + \gamma) = \alpha\beta + \alpha\gamma + \beta^2 + \beta\gamma \]

Combining these:
\[ ab + bc + ca = (\beta\gamma + \gamma\alpha + \alpha\beta) + (\beta\alpha + \beta^2 + \beta\gamma) + (\gamma^2 + \gamma\alpha + \alpha^2) \]
This simplifies to:
\[ = 2(\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha^2 + \beta^2 + \gamma^2) \]
Using \( \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \):
\[ = (-1)^2 - 2(2) = 1 - 4 = -3 \]
Thus,
\[ ab + bc + ca = 2(2) - 3 = 4 - 3 = 1 \]

### Step 3: Calculate the product of the roots
The product \( abc \) is given by:
\[ abc = (\beta + \gamma)(\gamma + \alpha)(\alpha + \beta) = \alpha\beta\gamma + \text{(other terms)} \]
Using Vieta's, we find:
\[ abc = -3 + 1 = -2 \]

### Step 4: Form the new polynomial
Now we have:
- \( a + b + c = -2 \)
- \( ab + bc + ca = 1 \)
- \( abc = -2 \)

The polynomial whose roots are \( a, b, c \) can be formed as:
\[ x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0 \]
Substituting the values:
\[ x^3 - (-2)x^2 + (1)x - (-2) = 0 \]
This simplifies to:
\[ x^3 + 2x^2 + x + 2 = 0 \]

### Final Answer
The equation whose roots are \( \beta + \gamma, \gamma + \alpha, \alpha + \beta \) is:
\[ \boxed{x^3 + 2x^2 + x + 2 = 0} \]