Math, asked by Suryavardhan1, 1 year ago

Please solve the question provided above:-

Attachments:

Answers

Answered by Mankuthemonkey01
19
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

Suryavardhan1: Thanks a lot bro for helping me out
Mankuthemonkey01: Welcome
Answered by Anonymous
11
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

Suryavardhan1: Thanks for answering the question @AhseFurieux
Anonymous: always welcome :)
Similar questions