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: ### Step 1: Start with the Given Equation You are given: \[ x - \frac{1}{x} = 4 \] ### Step 2: Square Both Sides of the Equation Squaring both sides will help eliminate the fraction and introduce \( x^2 \) and \( \frac{1}{x^2} \), which are present in the expression we need to prove. \[ \left( x - \frac{1}{x} \right)^2 = 4^2 \] Compute both sides: \[ x^2 - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 16 \] \[ x^2 - 2 + \frac{1}{x^2} = 16 \] ### Step 3: Simplify the Equation 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 Therefore, given that \( x - \frac{1}{x} = 4 \), it follows that: \[ x^{2} + \frac{1}{x^{2}} = 18 \]