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

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

(ii) $27x^3 +\dfrac{1}{27x^3}$

(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 $3x + \dfrac{1}{3x} = 3$,we get

$3^2 = 9x^2 + 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9 = 9x^2 + 2 + \dfrac{1}{9x^2}\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 9 - 2\\[1em] ⇒ 9x^2 + \dfrac{1}{9x^2} = 7$

Hence, the value of $9x^2 + \dfrac{1}{9x^2}$ is 7.

(ii) Using the formula,

[∵ (x + y)³ = x³ + 3xy(x + y) + y³]

So,

$\Big(3x + \dfrac{1}{3x}\Big)^3 = (3x)^3 + 3 \times 3x \times \dfrac{1}{3x}\Big(3x + \dfrac{1}{3x}\Big) + \Big(\dfrac{1}{3x}\Big)^3\\[1em] ⇒ \Big(3x + \dfrac{1}{3x}\Big)^3 = 27x^3 + 3\Big(3x + \dfrac{1}{3x}\Big) + \dfrac{1}{27x^3}$

Putting $3x + \dfrac{1}{3x} = 3$

$3^3 = 27x^3 + 3 \times 3 + \dfrac{1}{27x^3}\\[1em] ⇒ 27 = 27x^3 + 9 + \dfrac{1}{27x^3}\\[1em] ⇒ 27x^3 + \dfrac{1}{27x^3} = 27 - 9 \\[1em] ⇒ 27x^3 + \dfrac{1}{27x^3} = 18$

Hence, the value of $27x^3 + \dfrac{1}{27x^3}$ is 18.