Potential energy of a particle moving along x-axis is by
`U=(x^(3)/3-4x + 6)`.
here, U is in joule and x in metre. Find position of stable and unstable equilibrium.

Explanation

To find the positions of stable and unstable equilibrium for a particle moving along the x-axis with a given potential energy function \( U(x) = \frac{x^3}{3} - 4x + 6 \), we can follow these steps:

### Step 1: Find the Force
The force \( F \) acting on the particle is related to the potential energy \( U \) by the equation:
\[ F = -\frac{dU}{dx} \]
We need to differentiate \( U \) with respect to \( x \).

### Step 2: Differentiate the Potential Energy
Differentiate \( U(x) \):
\[ \frac{dU}{dx} = \frac{d}{dx}\left(\frac{x^3}{3} - 4x + 6\right) \]
Calculating the derivative:
\[ \frac{dU}{dx} = x^2 - 4 \]

### Step 3: Set the Force to Zero
To find the equilibrium positions, we set the force \( F \) to zero:
\[ F = -\frac{dU}{dx} = 0 \implies x^2 - 4 = 0 \]
Solving for \( x \):
\[ x^2 = 4 \implies x = \pm 2 \]

### Step 4: Determine Stability
To classify the equilibrium points as stable or unstable, we need to compute the second derivative of the potential energy:
\[ \frac{d^2U}{dx^2} = \frac{d}{dx}(x^2 - 4) = 2x \]

### Step 5: Evaluate the Second Derivative at Equilibrium Points
1. For \( x = 2 \):
\[ \frac{d^2U}{dx^2} \bigg|_{x=2} = 2(2) = 4 \quad (\text{positive, hence stable}) \]

2. For \( x = -2 \):
\[ \frac{d^2U}{dx^2} \bigg|_{x=-2} = 2(-2) = -4 \quad (\text{negative, hence unstable}) \]

### Conclusion
The positions of equilibrium are:
- **Stable equilibrium at \( x = 2 \) m**
- **Unstable equilibrium at \( x = -2 \) m**