If 3a + $\dfrac{1}{3a} = 2\sqrt{3}$, evaluate :

(i) 3a - $\dfrac{1}{3a}$

(ii) $9a^2 + \dfrac{1}{9a^2}$

(iii) $81a^4 + \dfrac{1}{81a^4}$

(i) Given: 3a + $\dfrac{1}{3a} = 2\sqrt{3}$

Squaring both sides, we get:

$\Rightarrow \Big(3a + \dfrac{1}{3a}\Big)^2 = (2\sqrt{3})^2\\[1em] \Rightarrow (3a)^2 + \Big(\dfrac{1}{3a}\Big)^2 + 2 \times 3a \times \dfrac{1}{3a} = 12\\[1em] \Rightarrow 9a^2 + \dfrac{1}{9a^2} + 2 = 12\\[1em] \Rightarrow 9a^2 + \dfrac{1}{9a^2} = 12 - 2\\[1em] \Rightarrow 9a^2 + \dfrac{1}{9a^2} = 10 …………………………(1)$

Using the formula: (a - b)² = a² + b² - 2ab

$\Rightarrow \Big(3a - \dfrac{1}{3a}\Big)^2 = (3a)^2 + \Big(\dfrac{1}{3a}\Big)^2 - 2 \times 3a \times \dfrac{1}{3a}\\[1em] \Rightarrow \Big(3a - \dfrac{1}{3a}\Big)^2 = 9a^2 + \dfrac{1}{9a^2} - 2\\[1em]$

Substituting the value of $9a^2 + \dfrac{1}{9a^2}$, we get:

$\Rightarrow \Big(3a - \dfrac{1}{3a}\Big)^2 = 10 - 2\\[1em] \Rightarrow \Big(3a - \dfrac{1}{3a}\Big)^2 = 8\\[1em] \Rightarrow 3a - \dfrac{1}{3a} = \sqrt{8}\\[1em] \Rightarrow 3a - \dfrac{1}{3a} = \pm 2\sqrt{2}$

Hence, 3a - $\dfrac{1}{3a} = \pm 2\sqrt{2}$.

(ii) From equation (1),

Hence, $9a^2 + \dfrac{1}{9a^2} = 10$.

(iii) Squaring equation (1), we get:

$\Rightarrow \Big(9a^2 + \dfrac{1}{9a^2}\Big)^2 = 10^2\\[1em] \Rightarrow (9a^2)^2 + \Big(\dfrac{1}{9a^2}\Big)^2 + 2 \times 9a^2 \times \Big(\dfrac{1}{9a^2}\Big) = 100\\[1em] \Rightarrow 81a^4 + \dfrac{1}{81a^4} + 2 = 100\\[1em] \Rightarrow 81a^4 + \dfrac{1}{81a^4} = 100 - 2\\[1em] \Rightarrow 81a^4 + \dfrac{1}{81a^4} = 98$

Hence, $81a^4 + \dfrac{1}{81a^4} = 98$.