f \( x-\frac{1}{x}=4 \), prove that \( x^{2}+\frac{1}{x^{2}}=18 \)

To prove that \( x^{2} + \frac{1}{x^{2}} = 18 \) given that \( x - \frac{1}{x} = 4 \), follow these steps: 1. **Start with the given equation:** \[ x - \frac{1}{x} = 4 \] 2. **Square both sides of the equation:** \[ \left( x - \frac{1}{x} \right)^{2} = 4^{2} \] 3. **Expand the left side using the identity \((a - b)^2 = a^2 - 2ab + b^2\):** \[ x^{2} - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} = 16 \] 4. **Simplify the middle term:** \[ x^{2} - 2 + \frac{1}{x^{2}} = 16 \] 5. **Combine like terms:** \[ x^{2} + \frac{1}{x^{2}} - 2 = 16 \] 6. **Add 2 to both sides to isolate \( x^{2} + \frac{1}{x^{2}} \):** \[ x^{2} + \frac{1}{x^{2}} = 16 + 2 \] \[ x^{2} + \frac{1}{x^{2}} = 18 \] **Conclusion:** Given that \( x - \frac{1}{x} = 4 \), it follows that \( x^{2} + \frac{1}{x^{2}} = 18 \).