Question

If x + 1/9x = 1, then find the value of 27x³ + 1/27x³ = ?​

Answer 1

$\large\underline{\sf{Solution-}}$

Given:

$\sf \longmapsto x + \dfrac{1}{9x} = 1$

Multiplying both sides by 3, we get:

$\sf \longmapsto 3 \bigg(x + \dfrac{1}{9x} \bigg) = 3 \times 1$

$\sf \longmapsto 3x + \dfrac{1}{3x} = 3$

Cubing on both sides, we get:

$\sf \longmapsto \bigg(3x + \dfrac{1}{3x} \bigg)^{3} = {3}^{3}$

$\sf \longmapsto {(3x)}^{3} + \bigg(\dfrac{1}{3x} \bigg)^{3} + \bigg(3 \times {(3x)}^{2} \times \dfrac{1}{3x} \bigg) + \bigg(3 \times 3x\times \bigg( \dfrac{1}{3x} \bigg)^{2} \bigg) = {3}^{3}$

$\sf \longmapsto 27{x}^{3} + \dfrac{1}{27x^{3}} + 9x + \dfrac{1}{x} = {3}^{3}$

$\sf \longmapsto 27{x}^{3} + \dfrac{1}{27x^{3}} + 9 \bigg(x + \dfrac{1}{9x} \bigg) = 27$

As we know that:

$\sf ,\red\longmapsto x + \dfrac{1}{9x} = 1$

We get:

$\sf \longmapsto 27{x}^{3} + \dfrac{1}{27x^{3}} + 9 \times 1 = 27$

$\sf \longmapsto 27{x}^{3} + \dfrac{1}{27x^{3}} + 9 = 27$

$\sf \longmapsto 27{x}^{3} + \dfrac{1}{27x^{3}} = 18 \: Ans.$