4. Complete the following. a) Simplify: i) \( \left(a-\frac{1}{a}\right)\left(a^{2}+1+\frac{1}{a^{2}}\right) \) ii) \( \left(a-\frac{1}{a}\right)^{2} \). b) Calculate the value of: i) \( a^{2}+\frac{1}{a^{2}} \), if \( a-\frac{1}{a}=3 \) ii) \( a^{3}-\frac{1}{a^{3}} \)

Certainly! Let's tackle each part of the problem step by step. --- ## **4. Complete the following.** ### **a) Simplify:** #### **i) \( \left(a - \frac{1}{a}\right)\left(a^{2} + 1 + \frac{1}{a^{2}}\right) \)** **Solution:** We can simplify the expression by expanding the multiplication: \[ \left(a - \frac{1}{a}\right)\left(a^{2} + 1 + \frac{1}{a^{2}}\right) \] Multiply each term in the first parenthesis by each term in the second parenthesis: \[ = a \cdot a^{2} + a \cdot 1 + a \cdot \frac{1}{a^{2}} - \frac{1}{a} \cdot a^{2} - \frac{1}{a} \cdot 1 - \frac{1}{a} \cdot \frac{1}{a^{2}} \] Simplify each term: \[ = a^{3} + a + \frac{a}{a^{2}} - \frac{a^{2}}{a} - \frac{1}{a} - \frac{1}{a^{3}} \] \[ = a^{3} + a + \frac{1}{a} - a - \frac{1}{a} - \frac{1}{a^{3}} \] Notice that some terms cancel out: \[ a - a = 0 \quad \text{and} \quad \frac{1}{a} - \frac{1}{a} = 0 \] So, we're left with: \[ = a^{3} - \frac{1}{a^{3}} \] **Final Answer:** \[ \left(a - \frac{1}{a}\right)\left(a^{2} + 1 + \frac{1}{a^{2}}\right) = a^{3} - \frac{1}{a^{3}} \] --- #### **ii) \( \left(a - \frac{1}{a}\right)^{2} \)** **Solution:** Expand the square using the formula \((x - y)^2 = x^2 - 2xy + y^2\): \[ \left(a - \frac{1}{a}\right)^{2} = a^{2} - 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^{2} \] Simplify each term: \[ = a^{2} - 2 \cdot 1 + \frac{1}{a^{2}} \] \[ = a^{2} + \frac{1}{a^{2}} - 2 \] **Final Answer:** \[ \left(a - \frac{1}{a}\right)^{2} = a^{2} + \frac{1}{a^{2}} - 2 \] --- ### **b) Calculate the value of:** #### **i) \( a^{2} + \frac{1}{a^{2}} \), if \( a - \frac{1}{a} = 3 \)** **Solution:** We are given: \[ a - \frac{1}{a} = 3 \] Square both sides to find \( a^{2} + \frac{1}{a^{2}} \): \[ \left(a - \frac{1}{a}\right)^{2} = 3^{2} \] Expand the left side: \[ a^{2} - 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^{2}} = 9 \] Simplify: \[ a^{2} - 2 + \frac{1}{a^{2}} = 9 \] Now, solve for \( a^{2} + \frac{1}{a^{2}} \): \[ a^{2} + \frac{1}{a^{2}} = 9 + 2 = 11 \] **Final Answer:** \[ a^{2} + \frac{1}{a^{2}} = 11 \] --- #### **ii) \( a^{3} - \frac{1}{a^{3}} \)** **Solution:** We can use the identity: \[ a^{3} - \frac{1}{a^{3}} = \left(a - \frac{1}{a}\right)\left(a^{2} + 1 + \frac{1}{a^{2}}\right) \] From part **b.i**, we have: \[ a - \frac{1}{a} = 3 \quad \text{and} \quad a^{2} + \frac{1}{a^{2}} = 11 \] Substitute these values into the identity: \[ a^{3} - \frac{1}{a^{3}} = 3 \cdot (11 + 1) = 3 \cdot 12 = 36 \] **Final Answer:** \[ a^{3} - \frac{1}{a^{3}} = 36 \] ---