If $2x -\dfrac{1}{2x} = 4$, find:

(i) $4x^2 +\dfrac{1}{4x^2}$

(ii) $8x^3 -\dfrac{1}{8x^3}$

(i) Using the formula,

[∵ (x - y)² = x² - 2xy + y²]

So,

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

Putting the value $2x - \dfrac{1}{2x} = 4$,we get

$4^2 = 4x^2 - 2 + \dfrac{1}{4x^2}\\[1em] ⇒ 16 = 4x^2 - 2 + \dfrac{1}{4x^2}\\[1em] ⇒ 4x^2 + \dfrac{1}{4x^2} = 16 + 2 \\[1em] ⇒ 4x^2 + \dfrac{1}{4x^2} = 18$

Hence, the value of $4x^2 + \dfrac{1}{4x^2}$ is 18.

(ii) Using the formula,

[∵ (x - y)³ = x³ - 3xy(x - y) - y³]

So,

$\Big(2x - \dfrac{1}{2x}\Big)^3 = (2x)^3 - 3 \times 2x \times \dfrac{1}{2x}\Big(2x - \dfrac{1}{2x}\Big) - \Big(\dfrac{1}{2x}\Big)^3\\[1em] ⇒ \Big(2x - \dfrac{1}{2x}\Big)^3 = 8x^3 - 3\Big(2x - \dfrac{1}{2x}\Big) - \dfrac{1}{8x^3}$

Putting $2x - \dfrac{1}{2x} = 4$

$4^3 = 8x^3 - 3 \times 4 - \dfrac{1}{8x^3}\\[1em] ⇒ 64 = 8x^3 - 12 - \dfrac{1}{8x^3}\\[1em] ⇒ 8x^3 - \dfrac{1}{8x^3} = 64 + 12 \\[1em] ⇒ 8x^3 - \dfrac{1}{8x^3} = 76$

Hence, the value of $8x^3 - \dfrac{1}{8x^3}$ = 76.