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 find the values of \( m \) such that the equations \( x^3 - mx^2 - 4 = 0 \) and \( x^3 + mx + 2 = 0 \) have one common root, we can follow these steps:

### Step 1: Let the common root be \( r \).
Assume that \( r \) is the common root of both equations.

### Step 2: Substitute \( r \) into both equations.
From the first equation:
\[ r^3 - mr^2 - 4 = 0 \quad \text{(1)} \]
From the second equation:
\[ r^3 + mr + 2 = 0 \quad \text{(2)} \]

### Step 3: Set the two equations equal.
Since both equations equal \( r^3 \), we can set them equal to each other:
\[ mr^2 + 4 = -mr - 2 \]

### Step 4: Rearranging the equation.
Rearranging gives:
\[ mr^2 + mr + 4 + 2 = 0 \]
\[ mr^2 + mr + 6 = 0 \quad \text{(3)} \]

### Step 5: Factor the quadratic equation.
Equation (3) is a quadratic in \( m \):
\[ m(r^2 + r) + 6 = 0 \]
From this, we can express \( m \):
\[ m = -\frac{6}{r^2 + r} \quad \text{(4)} \]

### Step 6: Find the values of \( m \) for specific roots.
To find specific values of \( m \), we can test for rational roots of the original equations. Let's try \( r = 1 \) and \( r = -2 \).

#### Case 1: \( r = 1 \)
Substituting \( r = 1 \) into equation (4):
\[ m = -\frac{6}{1^2 + 1} = -\frac{6}{2} = -3 \]

#### Case 2: \( r = -2 \)
Substituting \( r = -2 \) into equation (4):
\[ m = -\frac{6}{(-2)^2 + (-2)} = -\frac{6}{4 - 2} = -\frac{6}{2} = -3 \]

### Conclusion
Both cases yield the same value for \( m \):
\[ \boxed{-3} \]