If `alpha, beta` are zeroes of polynomial `6x^(2) +x-1`, then find the value of
(i) `alpha^(3) beta + alpha beta^(3)`, (ii) `alpha/beta + beta/alpha +2(1/alpha + 1/beta) + 3 alpha beta`

Answer

To solve the problem, we need to find the values of (i) \( \alpha^3 \beta + \alpha \beta^3 \) and (ii) \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta \), where \( \alpha \) and \( \beta \) are the roots of the polynomial \( 6x^2 + x - 1 \).

### Step 1: Find the roots \( \alpha \) and \( \beta \)

The polynomial is given as:
\[ p(x) = 6x^2 + x - 1 \]
Using the quadratic formula:
\[ \alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 6, b = 1, c = -1 \).

Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4 \cdot 6 \cdot (-1) = 1 + 24 = 25 \]
Now substituting into the quadratic formula:
\[ \alpha, \beta = \frac{-1 \pm \sqrt{25}}{2 \cdot 6} = \frac{-1 \pm 5}{12} \]
Calculating the roots:
1. \( \alpha = \frac{-1 + 5}{12} = \frac{4}{12} = \frac{1}{3} \)
2. \( \beta = \frac{-1 - 5}{12} = \frac{-6}{12} = -\frac{1}{2} \)

### Step 2: Calculate \( \alpha^3 \beta + \alpha \beta^3 \)

Using the values of \( \alpha \) and \( \beta \):
\[ \alpha^3 \beta + \alpha \beta^3 = \alpha \beta (\alpha^2 + \beta^2) \]
First, calculate \( \alpha \beta \):
\[ \alpha \beta = \frac{1}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{1}{6} \]

Next, calculate \( \alpha^2 + \beta^2 \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Calculating \( \alpha + \beta \):
\[ \alpha + \beta = \frac{1}{3} - \frac{1}{2} = \frac{2 - 3}{6} = -\frac{1}{6} \]
Now substituting:
\[ (\alpha + \beta)^2 = \left(-\frac{1}{6}\right)^2 = \frac{1}{36} \]
Now substituting into the equation:
\[ \alpha^2 + \beta^2 = \frac{1}{36} - 2\left(-\frac{1}{6}\right) = \frac{1}{36} + \frac{1}{3} = \frac{1}{36} + \frac{12}{36} = \frac{13}{36} \]
Now substituting back into \( \alpha^3 \beta + \alpha \beta^3 \):
\[ \alpha^3 \beta + \alpha \beta^3 = -\frac{1}{6} \cdot \frac{13}{36} = -\frac{13}{216} \]

### Step 3: Calculate \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta \)

Calculating \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} \):
\[ \frac{\alpha}{\beta} = \frac{\frac{1}{3}}{-\frac{1}{2}} = -\frac{2}{3}, \quad \frac{\beta}{\alpha} = \frac{-\frac{1}{2}}{\frac{1}{3}} = -\frac{3}{2} \]
Now adding:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{2}{3} - \frac{3}{2} = -\frac{4}{6} - \frac{9}{6} = -\frac{13}{6} \]

Calculating \( 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) \):
\[ \frac{1}{\alpha} = 3, \quad \frac{1}{\beta} = -2 \]
So:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = 3 - 2 = 1 \]
Thus:
\[ 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = 2 \cdot 1 = 2 \]

Calculating \( 3\alpha\beta \):
\[ 3\alpha\beta = 3 \cdot -\frac{1}{6} = -\frac{1}{2} \]

Now combining everything:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta = -\frac{13}{6} + 2 - \frac{1}{2} \]
Converting \( 2 \) and \( -\frac{1}{2} \) to sixths:
\[ 2 = \frac{12}{6}, \quad -\frac{1}{2} = -\frac{3}{6} \]
So:
\[ -\frac{13}{6} + \frac{12}{6} - \frac{3}{6} = -\frac{13 - 12 - 3}{6} = -\frac{4}{6} = -\frac{2}{3} \]

### Final Answers:
(i) \( \alpha^3 \beta + \alpha \beta^3 = -\frac{13}{216} \)

(ii) \( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta = -\frac{2}{3} \)