Question

If
$x + \frac{1}{x} = 7$
then find the value of
${x}^{3} + \frac{ 1 }{ {x}^{3} }$
​

Answer 1

$\bf{\large{\underline{\underline{Answer:-}}}}$

$\sf{x^3 + \dfrac{1}{x^3}= 322}$

$\bf{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x + \dfrac{1}{x} = 7}$

To find :- $\sf{x^3 + \dfrac{1}{x^3}}$

Solution :-

$x + \dfrac{1}{x} = 7$

Now on cubing on both the sides

${(x + \dfrac{1}{x}) }^{3} = {7}^{3}$

We know that, (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = x , y = 1/x

By substituting the values in the identity we have

${x}^{3} + \dfrac{1}{ {x}^{3}} + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 343$

${x}^{3} + \dfrac{1}{ {x}^{3} } + 3(x + \dfrac{1}{x}) = 343$

${x}^{3} + \dfrac{1}{ {x}^{3} } + 3(7) = 343$

[Since $\bf{x + \dfrac{1}{x} = 7}$ ]

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

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

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

$\Huge{\boxed{\sf{x^3 + \dfrac{1}{x^3} = 322}}}$

$\bf{\large{\underline{\underline{Identity \: Used:-}}}}$

(x + y)³ = x³ + y³ + 3xy(x + y)

$\bf{\large{\underline{\underline{Some \: important \: Identities:-}}}}$

• (x + y)² = x² + y² + 2xy

• (x - y)² = x² + y² - 2xy

• (x + y)(x - y) = x² - y²

• (x + a)(x + b) = x² + (a + b)x + ab

Answer 2

Given:

$x + \frac{1}{x} = 7$

To find: ${x}^{3} + \frac{ 1 }{ {x}^{3} }$

We know,

$\boxed{{(a + b)}^{3} = a^{3} + b^{3} + 3ab(a + b)}$

Here,

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

$\implies {(7)}^{3} = x^{3} + \dfrac{1}{x^{3}} + 3(7)$

$\implies 343 = x^{3} + \dfrac{1}{x^{3}} + 21$

$\implies x^{3} + \dfrac{1}{x^{3}} = 343 - 21 = 322$

$\mathfrak{\large{\underline{\underline{Answer}}}}$

$\boxed{x^{3} + \dfrac{1}{x^{3}} = 322}$