Suppose `alpha, beta, gamma ` are roots of `x^(3) + qx + r = 0 ` . If ` (alpha - beta)^(2) + (beta - gamma)^(2) + (gamma - alpha)^(2)` = - e, then q = ________

Answer

To solve the problem step by step, we will start with the given equation and the expression involving the roots.

### Step 1: Understanding the Roots
The roots of the polynomial \( x^3 + qx + r = 0 \) are \( \alpha, \beta, \gamma \). According to Vieta's formulas, we know:
- \( \alpha + \beta + \gamma = 0 \) (since there is no \( x^2 \) term)
- \( \alpha\beta + \beta\gamma + \gamma\alpha = q \)
- \( \alpha\beta\gamma = -r \)

### Step 2: Expanding the Given Expression
We need to evaluate the expression:
\[ (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 \]
Expanding each term, we have:
\[ (\alpha - \beta)^2 = \alpha^2 - 2\alpha\beta + \beta^2 \]
\[ (\beta - \gamma)^2 = \beta^2 - 2\beta\gamma + \gamma^2 \]
\[ (\gamma - \alpha)^2 = \gamma^2 - 2\gamma\alpha + \alpha^2 \]
Adding these together:
\[ (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = 2(\alpha^2 + \beta^2 + \gamma^2) - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]

### Step 3: Using the Relationship
We know from the problem statement that:
\[ (\alpha - \beta)^2 + (\beta - \gamma)^2 + (\gamma - \alpha)^2 = -e \]
Thus, we can set up the equation:
\[ 2(\alpha^2 + \beta^2 + \gamma^2) - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = -e \]

### Step 4: Expressing \( \alpha^2 + \beta^2 + \gamma^2 \)
Using the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Since \( \alpha + \beta + \gamma = 0 \):
\[ \alpha^2 + \beta^2 + \gamma^2 = 0 - 2q = -2q \]

### Step 5: Substituting Back
Substituting this back into our equation:
\[ 2(-2q) - 2q = -e \]
This simplifies to:
\[ -4q - 2q = -e \]
\[ -6q = -e \]
Thus, we find:
\[ q = \frac{e}{6} \]

### Conclusion
The value of \( q \) is:
\[ \boxed{\frac{e}{6}} \]