Let `f(x) = x^3 - 3x + b` and `g(x) = x^2 + bx - 3` where b is a real number. What is the sum of all possible values of b for which the equations f(x)= 0 and g(x) = 0 have a common root ?

Answer

To solve the problem, we need to find the sum of all possible values of \( b \) for which the equations \( f(x) = 0 \) and \( g(x) = 0 \) have a common root. Let's denote the common root as \( \alpha \).

1. **Set up the equations**:
We have:
\[ f(x) = x^3 - 3x + b \]
\[ g(x) = x^2 + bx - 3 \]
Since \( \alpha \) is a common root, it satisfies both equations:
\[ f(\alpha) = \alpha^3 - 3\alpha + b = 0 \quad \text{(1)} \]
\[ g(\alpha) = \alpha^2 + b\alpha - 3 = 0 \quad \text{(2)} \]

2. **Express \( b \) from both equations**:
From equation (1):
\[ b = -\alpha^3 + 3\alpha \quad \text{(3)} \]
From equation (2):
\[ b = \frac{3 - \alpha^2}{\alpha} \quad \text{(4)} \]

3. **Set the expressions for \( b \) equal**:
Equate equations (3) and (4):
\[ -\alpha^3 + 3\alpha = \frac{3 - \alpha^2}{\alpha} \]

4. **Multiply through by \( \alpha \) to eliminate the fraction**:
\[ -\alpha^4 + 3\alpha^2 = 3 - \alpha^2 \]
Rearranging gives:
\[ -\alpha^4 + 4\alpha^2 - 3 = 0 \]
Multiplying through by -1:
\[ \alpha^4 - 4\alpha^2 + 3 = 0 \]

5. **Let \( y = \alpha^2 \)**:
This transforms the equation into a quadratic:
\[ y^2 - 4y + 3 = 0 \]

6. **Factor the quadratic**:
\[ (y - 1)(y - 3) = 0 \]
Thus, \( y = 1 \) or \( y = 3 \).

7. **Find \( \alpha \)**:
Since \( y = \alpha^2 \):
- If \( y = 1 \), then \( \alpha^2 = 1 \) implies \( \alpha = 1 \) or \( \alpha = -1 \).
- If \( y = 3 \), then \( \alpha^2 = 3 \) implies \( \alpha = \sqrt{3} \) or \( \alpha = -\sqrt{3} \).

8. **Calculate corresponding values of \( b \)**:
- For \( \alpha = 1 \):
\[ b = -1^3 + 3 \cdot 1 = -1 + 3 = 2 \]
- For \( \alpha = -1 \):
\[ b = -(-1)^3 + 3 \cdot (-1) = 1 - 3 = -2 \]
- For \( \alpha = \sqrt{3} \):
\[ b = -(\sqrt{3})^3 + 3\sqrt{3} = -3\sqrt{3} + 3\sqrt{3} = 0 \]
- For \( \alpha = -\sqrt{3} \):
\[ b = -(-\sqrt{3})^3 + 3(-\sqrt{3}) = 3\sqrt{3} - 3\sqrt{3} = 0 \]

9. **Summing all possible values of \( b \)**:
The possible values of \( b \) are \( 2, -2, 0 \). Thus, the sum is:
\[ 2 + (-2) + 0 = 0 \]

**Final Answer**:
The sum of all possible values of \( b \) is \( \boxed{0} \).