Question

(i) Find the value of
$( {x}^{3} - \frac{1}{ {x}^{3} } )if \frac{x}{2} = \frac{1}{2x} + 1$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{x}{2}=\dfrac{1}{2x}+1}$

$\tt{\implies\,x=\dfrac{2}{2x}+2}$

$\tt{\implies\,x=\dfrac{1}{x}+2}$

$\tt{\implies\,x-\dfrac{1}{x}=2}$

$\tt{\implies\,\left(x-\dfrac{1}{x}\right)^3=\left(2\right)^3}$

$\tt{\implies\,\left(x\right)^3-\left(\dfrac{1}{x}\right)^3-3\cdot\,x\cdot\dfrac{1}{x}\cdot\left(x-\dfrac{1}{x}\right)=\left(2\right)^3}$

$\tt{\implies\,{x}^{3}-\dfrac{1}{{x}^{3}}-3\cdot2=8}$

$\tt{\implies\,{x}^{3}-\dfrac{1}{{x}^{3}}-6=8}$

$\tt{\implies\,{x}^{3}-\dfrac{1}{{x}^{3}}=14}$