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

Explanation

To solve the problem, we need to find the value of \( \alpha + \beta + \alpha \beta \) where \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( 3x^2 + 6x + 1 \).

### Step 1: Identify coefficients
The polynomial is in the form \( ax^2 + bx + c \), where:
- \( a = 3 \)
- \( b = 6 \)
- \( c = 1 \)

### Step 2: Calculate \( \alpha + \beta \)
Using the formula for the sum of the roots of a quadratic polynomial, we have:
\[ \alpha + \beta = -\frac{b}{a} \]
Substituting the values of \( b \) and \( a \):
\[ \alpha + \beta = -\frac{6}{3} = -2 \]

### Step 3: Calculate \( \alpha \beta \)
Using the formula for the product of the roots of a quadratic polynomial, we have:
\[ \alpha \beta = \frac{c}{a} \]
Substituting the values of \( c \) and \( a \):
\[ \alpha \beta = \frac{1}{3} \]

### Step 4: Calculate \( \alpha + \beta + \alpha \beta \)
Now, we can find \( \alpha + \beta + \alpha \beta \):
\[ \alpha + \beta + \alpha \beta = (-2) + \left(\frac{1}{3}\right) \]
To add these, we need a common denominator. The common denominator of 1 and 3 is 3:
\[ -2 = -\frac{6}{3} \]
So,
\[ \alpha + \beta + \alpha \beta = -\frac{6}{3} + \frac{1}{3} = -\frac{6 - 1}{3} = -\frac{5}{3} \]

### Final Answer
Thus, the value of \( \alpha + \beta + \alpha \beta \) is:
\[ \boxed{-\frac{5}{3}} \]