If the equations `x^(3) - mx^2 - 4 = 0 and x^(3) + mx + 2 = 0 .m in R` have one common root, then find the values of m.

Explanation

To solve the problem, we need to find the values of \( m \) for which the equations \( x^3 - mx^2 - 4 = 0 \) and \( x^3 + mx + 2 = 0 \) have one common root. Let's denote the common root as \( \alpha \).

### Step 1: Set up the equations
We have two equations:
1. \( x^3 - mx^2 - 4 = 0 \)
2. \( x^3 + mx + 2 = 0 \)

Since \( \alpha \) is a common root, we can substitute \( \alpha \) into both equations.

### Step 2: Substitute the common root
Substituting \( \alpha \) into the first equation:
\[ \alpha^3 - m\alpha^2 - 4 = 0 \quad \text{(1)} \]

Substituting \( \alpha \) into the second equation:
\[ \alpha^3 + m\alpha + 2 = 0 \quad \text{(2)} \]

### Step 3: Eliminate \( \alpha^3 \)
From equation (1), we can express \( \alpha^3 \):
\[ \alpha^3 = m\alpha^2 + 4 \]

Now, we can substitute this expression for \( \alpha^3 \) into equation (2):
\[ m\alpha^2 + 4 + m\alpha + 2 = 0 \]
This simplifies to:
\[ m\alpha^2 + m\alpha + 6 = 0 \]

### Step 4: Factor the equation
We can factor this equation:
\[ m(\alpha^2 + \alpha) + 6 = 0 \]

### Step 5: Solve for \( m \)
Rearranging gives:
\[ m(\alpha^2 + \alpha) = -6 \]
Thus,
\[ m = \frac{-6}{\alpha^2 + \alpha} \]

### Step 6: Find conditions for \( m \)
For \( m \) to be real, \( \alpha^2 + \alpha \) must not be zero. Therefore, we need to find the values of \( \alpha \) such that \( \alpha^2 + \alpha \neq 0 \).

### Step 7: Solve \( \alpha^2 + \alpha = 0 \)
Factoring gives:
\[ \alpha(\alpha + 1) = 0 \]
This means \( \alpha = 0 \) or \( \alpha = -1 \).

### Step 8: Calculate \( m \) for valid \( \alpha \)
1. If \( \alpha = 0 \):
\[ m = \frac{-6}{0^2 + 0} \quad \text{(undefined)} \]

2. If \( \alpha = -1 \):
\[ m = \frac{-6}{(-1)^2 + (-1)} = \frac{-6}{1 - 1} \quad \text{(undefined)} \]

### Step 9: Check for other values of \( \alpha \)
Since \( \alpha \) cannot be \( 0 \) or \( -1 \), we can choose other values. Let's check \( \alpha = 1 \) and \( \alpha = -2 \):

1. If \( \alpha = 1 \):
\[ m = \frac{-6}{1^2 + 1} = \frac{-6}{2} = -3 \]

2. If \( \alpha = -2 \):
\[ m = \frac{-6}{(-2)^2 + (-2)} = \frac{-6}{4 - 2} = \frac{-6}{2} = -3 \]

### Conclusion
In both cases, we find that:
\[ m = -3 \]

Thus, the value of \( m \) for which the equations have one common root is:
\[ \boxed{-3} \]