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) \( a^{3}-\frac{1}{a^{3}} \)

To simplify the given expressions: a) i) For \( \left(a-\frac{1}{a}\right)\left(a^{2}+1+\frac{1}{a^{2}}\right) \), let's first expand it: \[ = a \cdot \left(a^{2}+1+\frac{1}{a^{2}}\right) - \frac{1}{a} \cdot \left(a^{2}+1+\frac{1}{a^{2}}\right) \] \[ = a^3 + a - 1 - \frac{1}{a} + \frac{1}{a^3} \] Combine the terms to get: \[ = a^3 + \frac{1}{a^3} + a - 1 - \frac{1}{a} \] a) ii) For \( \left(a-\frac{1}{a}\right)^{2} \): \[ = a^2 - 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^{2} \] \[ = a^2 - 2 + \frac{1}{a^2} \] So, the final result is: \[ = a^2 + \frac{1}{a^2} - 2 \] b) i) Given \( a-\frac{1}{a}=3 \), we can use the identity for \( a^2+\frac{1}{a^2} \): \[ a^2+\frac{1}{a^2} = \left(a-\frac{1}{a}\right)^{2} + 2 \] \[ = 3^2 + 2 = 9 + 2 = 11 \] b) ii) To find \( a^{3}-\frac{1}{a^{3}} \): Use the identity: \[ a^{3}-\frac{1}{a^{3}} = \left(a-\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}+1\right) \] From a), \( a^2+\frac{1}{a^2} = 11 \): \[ = 3(11+1) = 3 \cdot 12 = 36 \] So, the simplified expressions and calculations yield: i) \( a^3 + \frac{1}{a^3} + a - 1 - \frac{1}{a} \) ii) \( a^2 + \frac{1}{a^2} - 2 \) iii) \( 11 \) iv) \( 36 \)