If `alpha,beta` are zeroes of polynomial f(x)`=x^(2)-p(x+1)-c`, then find the value of `(alpha+1)(beta+1)`.

Explanation

To solve the problem, we need to find the value of \((\alpha + 1)(\beta + 1)\) given that \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(f(x) = x^2 - p(x + 1) - c\).

### Step-by-step Solution:

1. **Identify the Polynomial**:
The polynomial is given as:
\[ f(x) = x^2 - p(x + 1) - c \]
We can rewrite this as:
\[ f(x) = x^2 - px - p - c \]

2. **Identify the Coefficients**:
From the polynomial \(f(x) = x^2 - px - (p + c)\), we can identify the coefficients:
- Coefficient of \(x^2\) is \(1\)
- Coefficient of \(x\) is \(-p\)
- Constant term is \(-(p + c)\)

3. **Use Vieta's Formulas**:
According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta\) is given by:
\[ \alpha + \beta = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} = -\frac{-p}{1} = p \]
- The product of the roots \(\alpha \beta\) is given by:
\[ \alpha \beta = \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{-(p + c)}{1} = - (p + c) \]

4. **Calculate \((\alpha + 1)(\beta + 1)\)**:
We can expand \((\alpha + 1)(\beta + 1)\):
\[ (\alpha + 1)(\beta + 1) = \alpha \beta + \alpha + \beta + 1 \]
Substituting the values from Vieta's formulas:
\[ = \alpha \beta + (\alpha + \beta) + 1 \]
\[ = - (p + c) + p + 1 \]
\[ = -p - c + p + 1 \]
\[ = -c + 1 \]

5. **Final Result**:
Therefore, the value of \((\alpha + 1)(\beta + 1)\) is:
\[ = 1 - c \]

### Conclusion:
The final answer is:
\[ (\alpha + 1)(\beta + 1) = 1 - c \]