If `alpha` and `beta` are zeroes of the quadratic polynomial f(x)`=3x^(2)-5x-2`, then find the value of `((alpha^(2))/(beta)+(beta^(2))/(alpha))+6(alpha+1)(beta+1)`.

Explanation

To solve the problem, we need to find the value of the expression:

\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} + 6(\alpha + 1)(\beta + 1) \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic polynomial \(f(x) = 3x^2 - 5x - 2\).

### Step 1: Find the sum and product of the roots
For a quadratic polynomial of the form \(ax^2 + bx + c\), the sum and product of the roots can be calculated using the formulas:
- Sum of roots (\(\alpha + \beta\)) = \(-\frac{b}{a}\)
- Product of roots (\(\alpha \beta\)) = \(\frac{c}{a}\)

Here, \(a = 3\), \(b = -5\), and \(c = -2\).

Calculating the sum:
\[ \alpha + \beta = -\frac{-5}{3} = \frac{5}{3} \]

Calculating the product:
\[ \alpha \beta = \frac{-2}{3} \]

### Step 2: Substitute the values into the expression
Now we can rewrite the expression \(\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}\) using the sum and product of the roots.

Using the identity:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta} \]

We can find \(\alpha^3 + \beta^3\) using the formula:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We know:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values:
\[ \alpha^2 + \beta^2 = \left(\frac{5}{3}\right)^2 - 2\left(-\frac{2}{3}\right) = \frac{25}{9} + \frac{4}{3} = \frac{25}{9} + \frac{12}{9} = \frac{37}{9} \]

Now substituting back:
\[ \alpha^3 + \beta^3 = \left(\frac{5}{3}\right)\left(\frac{37}{9} + \frac{2}{3}\right) \]
Calculating \(\frac{2}{3}\) in terms of ninths:
\[ \frac{2}{3} = \frac{6}{9} \]
So,
\[ \alpha^3 + \beta^3 = \left(\frac{5}{3}\right)\left(\frac{37 + 6}{9}\right) = \left(\frac{5}{3}\right)\left(\frac{43}{9}\right) = \frac{215}{27} \]

Now substituting into the expression:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\frac{215}{27}}{-\frac{2}{3}} = \frac{215 \times 3}{27 \times -2} = -\frac{645}{54} \]

### Step 3: Calculate \(6(\alpha + 1)(\beta + 1)\)
Now we calculate \(6(\alpha + 1)(\beta + 1)\):
\[ (\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 = -\frac{2}{3} + \frac{5}{3} + 1 = -\frac{2}{3} + \frac{5}{3} + \frac{3}{3} = \frac{6}{3} = 2 \]
Thus,
\[ 6(\alpha + 1)(\beta + 1) = 6 \times 2 = 12 \]

### Step 4: Combine the results
Now, we combine the results:
\[ \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} + 6(\alpha + 1)(\beta + 1) = -\frac{645}{54} + 12 \]
Converting 12 into a fraction with a denominator of 54:
\[ 12 = \frac{648}{54} \]
So,
\[ -\frac{645}{54} + \frac{648}{54} = \frac{3}{54} = \frac{1}{18} \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{\frac{1}{18}} \]