Question

$\frac{x^2 + 1 }{x} = 7 , Find : \\1) x+ \frac{1}{x} \\\\ 2) x^3+\frac{1}{x^3}$

Answer 1

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

$\rm :\longmapsto\: \dfrac{ {x}^{2} + 1}{x} = 7$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:1) \: \: \: \: \: x + \dfrac{1}{x}$

$\rm :\longmapsto\:2) \: \: \: \: \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

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

Given that

$\rm :\longmapsto\: \dfrac{ {x}^{2} + 1}{x} = 7$

$\rm :\longmapsto\: \dfrac{ {x}^{2}}{x} + \dfrac{1}{x} = 7$

$\bf\implies \: \: x+ \: \dfrac{1}{x} = 7$

Now, On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{3} = {7}^{3}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg(x + \dfrac{1}{x} \bigg) = 343$

$\: \: \: \: \: \: \red{\bigg \{ \because \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \bigg \}}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 7 = 343$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: x + \dfrac{1}{x} = 7\bigg \}}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 21 = 343$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 343 - 21$

$\bf :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 322$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)