Question

If
$x + 1 \x = 5 then the vaule of x {3} + 1 \x {3}$
​

Answer 1

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

$\: \: \: \: \: \: \bull \: \: \sf \: x + \dfrac{1}{x} = 5$

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

$\: \: \: \: \: \: \bull \: \: \sf \: {x}^{3} + \dfrac{1}{ {x}^{3} }$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \sf \: {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}$

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

Given that

$\rm :\longmapsto\:x + \dfrac{1}{x} = 5$

On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{x} \bigg) }^{3} = {(5)}^{3}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \bigg(x + \dfrac{1}{x} \bigg) = 125$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 5 = 125$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } +15 = 125$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 125 - 15$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {x}^{3} } = 110$

Additional Information :-

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)