Let `alpha` satisfy the equation `x ^(3) +3x ^(2) +4x+5=0 and beta` satisfy the equatin `x ^(3) -3x ^(2)+4x-5=0,alpha, beta in R, ` then `alpha + beta =`

Answer

To solve the problem, we need to find the value of \( \alpha + \beta \) where \( \alpha \) satisfies the equation \( x^3 + 3x^2 + 4x + 5 = 0 \) and \( \beta \) satisfies the equation \( x^3 - 3x^2 + 4x - 5 = 0 \).

### Step-by-Step Solution:

1. **Write the equations for \( \alpha \) and \( \beta \)**:
- For \( \alpha \):
\[ \alpha^3 + 3\alpha^2 + 4\alpha + 5 = 0 \tag{1} \]
- For \( \beta \):
\[ \beta^3 - 3\beta^2 + 4\beta - 5 = 0 \tag{2} \]

2. **Add the two equations**:
\[ (\alpha^3 + 3\alpha^2 + 4\alpha + 5) + (\beta^3 - 3\beta^2 + 4\beta - 5) = 0 \]
This simplifies to:
\[ \alpha^3 + \beta^3 + 3\alpha^2 - 3\beta^2 + 4\alpha + 4\beta + 0 = 0 \]
Thus, we can rearrange it as:
\[ \alpha^3 + \beta^3 + 3(\alpha^2 - \beta^2) + 4(\alpha + \beta) = 0 \]

3. **Factor \( \alpha^2 - \beta^2 \)**:
Using the identity \( a^2 - b^2 = (a-b)(a+b) \):
\[ \alpha^2 - \beta^2 = (\alpha - \beta)(\alpha + \beta) \]
Therefore, we can rewrite the equation as:
\[ \alpha^3 + \beta^3 + 3(\alpha - \beta)(\alpha + \beta) + 4(\alpha + \beta) = 0 \]

4. **Use the identity for \( \alpha^3 + \beta^3 \)**:
The identity for the sum of cubes is:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
Substituting this into our equation gives:
\[ (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) + 3(\alpha - \beta)(\alpha + \beta) + 4(\alpha + \beta) = 0 \]

5. **Factor out \( \alpha + \beta \)**:
Factoring out \( \alpha + \beta \):
\[ (\alpha + \beta)\left(\alpha^2 - \alpha\beta + \beta^2 + 3(\alpha - \beta) + 4\right) = 0 \]
This gives us two cases:
- Case 1: \( \alpha + \beta = 0 \)
- Case 2: \( \alpha^2 - \alpha\beta + \beta^2 + 3(\alpha - \beta) + 4 = 0 \)

6. **Conclusion**:
Since we are asked to find \( \alpha + \beta \), we can conclude from Case 1:
\[ \alpha + \beta = 0 \]

### Final Answer:
\[ \alpha + \beta = 0 \]