Question

Please solve the question provided above:-

Answer 1

Given

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

To find,

${x}^{3} + \frac{1}{ {x}^{3} }$

Solution :-

Cube both sides in the equation
$(x + \frac{1}{x} ) {}^{3} = (7) {}^{3}$

Now, we know that,

(a + b)³ = a³ + b³ + 3ab(a + b)

So,
$(x + \frac{1}{x{}^{} } )^{3} = {7}^{3}$

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

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

$since, \: x + \frac{1}{x} = 7$

=>
${x}^{3} + \frac{1}{ {x}^{3} } + 21 = 343 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 343 - 21 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 322$

Answer :- 322

Answer 2

Given :
$x + \frac{1}{x} = 7$
To Find :

${x}^{3} + \frac{1}{ {x}^{3} }$


Solution :

${(x + \frac{1}{x} )}^{3} = {x}^{3} + ({\frac{1}{x}})^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ \\ {7}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3(7) \\ \\ 343 = {x}^{3} + \frac{1}{ {x}^{3} } + 21 \\ \\ 343 - 21 = {x}^{3} + \frac{1}{ {x}^{3} } \\ \\ = 322$

The required Answer is 322