If ` alpha, beta, gamma ` be the roots of the equation ` x(1+x^(2))+x^(2)(6+x)+2=0` then match the entries of column-I with those of column-II.
`{:("column-I" , " Column -II"),("(A) " alpha^(-1)+beta^(-1)+gamma^(-1) " is equal to " , "(p) 8"),("(B)" alpha^(2)+beta^(2)+gamma^(2)" equals" , "(q)" -1/2),("(C)"(alpha^(-1)+beta^(-1)+gamma^(-1))-(alpha+beta+gamma) " is equal to " , "(r) -1"),("(D)"[alpha^(-1)+beta^(-1)+gamma^(-1)] " equals where [.] denotes the greatest integer equal to " , "(t)" 5/2):}`

Answer

To solve the equation \( x(1+x^{2}) + x^{2}(6+x) + 2 = 0 \) and find the relationships between the roots \( \alpha, \beta, \gamma \), we will follow these steps:

### Step 1: Simplify the Equation
Start with the given equation:
\[ x(1+x^{2}) + x^{2}(6+x) + 2 = 0 \]
Distributing the terms:
\[ x + x^{3} + 6x^{2} + x^{3} + 2 = 0 \]
Combining like terms:
\[ 2x^{3} + 6x^{2} + x + 2 = 0 \]

### Step 2: Identify Coefficients
This is a cubic equation of the form \( ax^3 + bx^2 + cx + d = 0 \) where:
- \( a = 2 \)
- \( b = 6 \)
- \( c = 1 \)
- \( d = 2 \)

### Step 3: Use Vieta's Formulas
From Vieta's formulas, we can find:
1. The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{6}{2} = -3 \)
2. The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{1}{2} \)
3. The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{2}{2} = -1 \)

### Step 4: Calculate \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \)
Using the formula:
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} \]
Substituting the values:
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} = \frac{\frac{1}{2}}{-1} = -\frac{1}{2} \]
Thus, \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = -\frac{1}{2} \) which corresponds to entry \( (q) \).

### Step 5: Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the known values:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-3)^2 - 2 \times \frac{1}{2} = 9 - 1 = 8 \]
Thus, \( \alpha^2 + \beta^2 + \gamma^2 = 8 \) which corresponds to entry \( (p) \).

### Step 6: Calculate \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} - (\alpha + \beta + \gamma) \)
We already found \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = -\frac{1}{2} \) and \( \alpha + \beta + \gamma = -3 \):
\[ \alpha^{-1} + \beta^{-1} + \gamma^{-1} - (\alpha + \beta + \gamma) = -\frac{1}{2} - (-3) = -\frac{1}{2} + 3 = \frac{5}{2} \]
Thus, this corresponds to entry \( (t) \).

### Step 7: Calculate the Greatest Integer of \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} \)
We already found \( \alpha^{-1} + \beta^{-1} + \gamma^{-1} = -\frac{1}{2} \).
The greatest integer function \( \lfloor -\frac{1}{2} \rfloor = -1 \).
Thus, this corresponds to entry \( (r) \).

### Summary of Results
- \( (A) \) corresponds to \( (q) \)
- \( (B) \) corresponds to \( (p) \)
- \( (C) \) corresponds to \( (t) \)
- \( (D) \) corresponds to \( (r) \)