Question

Need helps of brainly experts:

​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

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

Multiplying both sides by 3, we get:

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

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

Cubing both sides, we get:

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

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

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

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

As we know that:

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

We get:

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

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

$\sf \implies 27{x}^{3} + \dfrac{1}{27x^{3}} = 18$

Which is our required answer.

$\texttt{\textsf{\large{\underline{Answer}:}}}$

• 27x³ + 1/27x³ = 18

$\texttt{\textsf{\large{\underline{More Identities To Know}:}}}$

1. (a + b)² = a² + 2ab + b²
2. (a - b)² = a² - 2ab + b²
3. a² - b² = (a + b)(a - b)
4. (a + b)³ = a³ + 3ab(a + b) + b³
5. (a - b)³ = a³ - 3ab(a - b) - b³
6. a³ + b³ = (a + b)(a² - ab + b²)
7. a³ - b³ = (a - b)(a² + ab + b²)