Question

help me to solve this question

Answer 1

Given : (x - 1/x) = 3.

On cubing both sides, we get

$=> (x - \frac{1}{x})^3 = (3)^3$

$=> x^3 - \frac{1}{x^3}-3(x - \frac{1}{x}) = 27$

$=> x^3 - \frac{1}{x^3} - 3(3) = 27$

$=> x^3 - \frac{1}{x^3} - 9 = 27$

$= > x^3 - \frac{1}{x^3} = 27 + 9$

$=> x^3 - \frac{1}{x^3} = 36$

Hope this helps!

Answer 2

$\textbf{Hey mate! here's your answer! }$

$\textbf{ given }$ =

$x - \frac{1}{x} = 3$

$\textbf{ To find }$ =

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

$\textbf{ So, Answer }$ =

BY USING THE IDENTITY,

(x-y)³ = x³ - y³ - 3xy( x-y ),

We make the cube of the given expression,

${( x - \frac{1}{x} )}^{3} = ({x})^{3} - ({\frac{1}{x})}^{3} - 3(x)( \frac{1}{x} )(x - \frac{1}{x} )$


$= > ({3})^{3} = {x}^{3} - \frac{1}{{x}^{3} } - 3(3)$


$= > 27 = {x}^{3} - \frac{1}{ {x}^{3} } - 9$


$= > {x}^{3} - \frac{1}{ {x}^{3} } = 27 + 9$


$= > {x}^{3} - \frac{1}{ {x}^{3} } = 36$



$\textbf{ hope you understood}$