If $3x -\dfrac{1}{3x}= 5$, find:

(i) $9x^2 +\dfrac{1}{9x^2}$

(ii) $81x^4 +\dfrac{1}{81x^4}$

(i) Using the formula,

[∵ (x - y)² = x² - 2xy + y²]

So,

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

Putting the value $3x - \dfrac{1}{3x} = 5$,we get

$5^2 = 9x^2 - 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 25 = 9x^2 - 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 25 + 2 \\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 27$

Hence, the value of $9x^2 + \dfrac{1}{9x^2}$ is 27.

(ii) Using the formula,

[∵ (x + y)² = x² + 2xy + y²]

So,

$\Big(9x^2 + \dfrac{1}{9x^2}\Big)^2 = (9x^2)^2 + 2 \times 9x^2 \times \dfrac{1}{9x^2} + \Big(\dfrac{1}{9x^2}\Big)^2\\[1em] ⇒ \Big(9x^2 + \dfrac{1}{9x^2}\Big)^2 = 81x^4 + 2 + \dfrac{1}{81x^4}$

Putting the value $9x^2 + \dfrac{1}{9x^2} = 27$,we get

$27^2 = 81x^4 + 2 + \dfrac{1}{81x^4}\\[1em] ⇒ 729 = 81x^4 + 2 + \dfrac{1}{81x^4}\\[1em] ⇒ 81x^4 + \dfrac{1}{81x^4} = 729 - 2 \\[1em] ⇒ 81x^4 + \dfrac{1}{81x^4} = 727$

Hence, the value of $81x^4 + \dfrac{1}{81x^4}$ is 727.