(b) If \( x-\frac{1}{x}=2 \), determine, without solving the given equation, the value(s) of the following expressions. \( \begin{array}{lll}\text { (1) } x^{2}+\frac{1}{x^{2}} & \text { (2) } x+\frac{1}{x} & \text { (3) } x^{3}-\frac{1}{x^{3}}\end{array} \)

To find \( x^2 + \frac{1}{x^2} \), we can start by squaring both sides of the original equation \( x - \frac{1}{x} = 2 \). This gives: \[ (x - \frac{1}{x})^2 = 2^2 \] Expanding this, we get: \[ x^2 - 2 + \frac{1}{x^2} = 4 \] So, \[ x^2 + \frac{1}{x^2} = 4 + 2 = 6 \] Next, we can find \( x + \frac{1}{x} \) by rearranging the equation. From \( x - \frac{1}{x} = 2 \), we can add \( 2 \frac{1}{x} \) to both sides: \[ x + \frac{1}{x} = 2 + 2 \frac{1}{x} \] To find \( 2 \frac{1}{x} \), rearranging gives: Multiplying both sides of the original equation \( x - \frac{1}{x} = 2 \) by \( \frac{1}{x} \), we find \( 2 \frac{1}{x} = 2 + (x - 2) \). This is a step towards finding \( x + \frac{1}{x} \), but we can also use the relationship \( x + \frac{1}{x} = (x - \frac{1}{x}) + 2 \): Thus: \[ x + \frac{1}{x} = 2 + 2 = 4 \] Finally, to find \( x^3 - \frac{1}{x^3} \), we can use the identity: \[ x^3 - \frac{1}{x^3} = (x - \frac{1}{x}) \left( x^2 + 1 + \frac{1}{x^2} \right) \] Since we have found \( x - \frac{1}{x} = 2 \) and \( x^2 + \frac{1}{x^2} = 6 \): Now substituting: \[ x^3 - \frac{1}{x^3} = 2 \left( x^2 + 1 + \frac{1}{x^2} \right) \] Calculating \( x^2 + 1 + \frac{1}{x^2} \): Since we already have \( x^2 + \frac{1}{x^2} = 6\), adding 1 gives: \[ x^2 + 1 + \frac{1}{x^2} = 6 + 1 = 7 \] Thus: \[ x^3 - \frac{1}{x^3} = 2 \times 7 = 14 \] In summary: 1. \( x^2 + \frac{1}{x^2} = 6 \) 2. \( x + \frac{1}{x} = 4 \) 3. \( x^3 - \frac{1}{x^3} = 14 \)