If $a = \dfrac{1}{a - 5}$, where $a \neq 5 \text{ and }a \neq 0$, find the values of:

(i) $\Big(a - \dfrac{1}{a}\Big)$

(ii) $\Big(a + \dfrac{1}{a}\Big)$

(iii) $\Big(a^2 - \dfrac{1}{a^2}\Big)$

(iv) $\Big(a^2 + \dfrac{1}{a^2}\Big)$

(i) Given,

$a = \dfrac{1}{a - 5}$

$\Rightarrow a = \dfrac{1}{a - 5} \\[1em] \Rightarrow a - 5 = \dfrac{1}{a} \\[1em] \Rightarrow a - \dfrac{1}{a} = 5$

Hence, $\Big(a - \dfrac{1}{a}\Big) = 5$

(ii) From part (i),

$a - \dfrac{1}{a} = 5$

Using identity,

$\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (5)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 25\\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 29 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}$

Hence, $a + \dfrac{1}{a} = \pm \sqrt{29}$

(iii) From (i) and (ii),

$\Rightarrow a - \dfrac{1}{a} = 5 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm \sqrt{29}$

Using identity,

$\Big(a + \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a}\Big) = \Big( a^2 - \dfrac{1}{a^2}\Big)$

$\Rightarrow (\pm \sqrt{29}) \times (5) = \Big( a^2 - \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow \Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 5\sqrt{29} \\[1em]$

Hence, $\Big( a^2 - \dfrac{1}{a^2}\Big) = \pm 5\sqrt{29}$

(iv) From (i) and (ii),

$\Rightarrow a - \dfrac{1}{a} = 5 \\[1em] \Rightarrow a + \dfrac{1}{a} = \pm\sqrt{29}$

Using identity,

$\Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big)$

$\Rightarrow (\pm \sqrt{29})^2 + (5)^2 = 2\Big( a^2 + \dfrac{1}{a^2}\Big) \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 29 + 25 \\[1em] \Rightarrow 2\Big( a^2 + \dfrac{1}{a^2}\Big) = 54 \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = \dfrac{54}{2} \\[1em] \Rightarrow \Big( a^2 + \dfrac{1}{a^2}\Big) = 27 \\[1em]$

Hence, $\Big(a^2 + \dfrac{1}{a^2}\Big) = 27$.