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 \( \alpha \).

### Step 1: Set up the equations
We have:
\[ f(x) = x^3 - 3x + b = 0 \]
\[ g(x) = x^2 + bx - 3 = 0 \]
Substituting \( \alpha \) into both equations gives us:
\[ \alpha^3 - 3\alpha + b = 0 \quad \text{(1)} \]
\[ \alpha^2 + b\alpha - 3 = 0 \quad \text{(2)} \]

### Step 2: Solve for \( b \) in terms of \( \alpha \)
From equation (2), we can express \( b \) in terms of \( \alpha \):
\[ b\alpha = 3 - \alpha^2 \]
\[ b = \frac{3 - \alpha^2}{\alpha} \quad \text{(3)} \]

### Step 3: Substitute \( b \) into equation (1)
Now, substitute \( b \) from equation (3) into equation (1):
\[ \alpha^3 - 3\alpha + \frac{3 - \alpha^2}{\alpha} = 0 \]
Multiplying through by \( \alpha \) to eliminate the fraction gives:
\[ \alpha^4 - 3\alpha^2 + 3 - \alpha^2 = 0 \]
This simplifies to:
\[ \alpha^4 - 4\alpha^2 + 3 = 0 \]

### Step 4: Let \( y = \alpha^2 \)
Let \( y = \alpha^2 \). Then the equation becomes:
\[ y^2 - 4y + 3 = 0 \]

### Step 5: Factor the quadratic equation
Factoring the quadratic:
\[ (y - 1)(y - 3) = 0 \]
Thus, we have:
\[ y = 1 \quad \text{or} \quad y = 3 \]

### Step 6: Find \( \alpha \) values
Returning to \( \alpha \):
1. If \( y = 1 \), then \( \alpha^2 = 1 \) implies \( \alpha = 1 \) or \( \alpha = -1 \).
2. If \( y = 3 \), then \( \alpha^2 = 3 \) implies \( \alpha = \sqrt{3} \) or \( \alpha = -\sqrt{3} \).

### Step 7: Find corresponding \( b \) values
Now we will find \( b \) for each \( \alpha \):
1. For \( \alpha = 1 \):
\[ b = \frac{3 - 1^2}{1} = \frac{3 - 1}{1} = 2 \]
2. For \( \alpha = -1 \):
\[ b = \frac{3 - (-1)^2}{-1} = \frac{3 - 1}{-1} = -2 \]
3. For \( \alpha = \sqrt{3} \):
\[ b = \frac{3 - (\sqrt{3})^2}{\sqrt{3}} = \frac{3 - 3}{\sqrt{3}} = 0 \]
4. For \( \alpha = -\sqrt{3} \):
\[ b = \frac{3 - (-\sqrt{3})^2}{-\sqrt{3}} = \frac{3 - 3}{-\sqrt{3}} = 0 \]

### Step 8: Sum of all possible values of \( b \)
The possible values of \( b \) are \( 2, -2, \) and \( 0 \). Thus, the sum is:
\[ 2 + (-2) + 0 = 0 \]

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