Question

x=4+√15.findx^3-1/x^3​

Answer 1

Given -

$x = 4 + \sqrt{15}$

To find -

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

Solution -

$x = 4 + \sqrt{15}$

$\frac{1}{x} = \frac{1}{4 + \sqrt{15} }$

By rationalising the denominator we will get -

$\frac{1}{x} = \frac{4 - \sqrt{15} }{(4 + \sqrt{15})(4 - \sqrt{15} ) }$

$\frac{1}{x} = \frac{4 - \sqrt{15} }{ {(4)}^{2} - {( \sqrt{15}) }^{2} }$

$\frac{1}{x} = \frac{4 - \sqrt{15} }{16 - 15}$

$\frac{1}{x} = \frac{4 - \sqrt{15} }{1}$

$\frac{1}{x} = 4 - \sqrt{15}$

Now,

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

Putting the values we will get -

$= {(4 + \sqrt{15} )}^{3} - {(4 - \sqrt{15}) }^{3}$

$= {(4)}^{3} + {( \sqrt{15} )}^{3} + 3 {(4)}^{2} \times \sqrt{15} + 3 \times 4 {( \sqrt{15}) }^{2} -[ {(4)}^{3} - {( \sqrt{15}) }^{3} - 3( {4)}^{2} \sqrt{15} + 3 \times 4 \times \sqrt{15} {}^{2}]$

$= {(4)}^{3} + {( \sqrt{15} )}^{3} + 3 {(4)}^{2} \sqrt{15} + 3 \times 4 \times {( \sqrt{15} )}^{2} - {(4)}^{3} + {( \sqrt{15}) }^{3} + 3 {(4)}^{2} \sqrt{15} - 3 \times 4 \times {( \sqrt{15}) }^{2}$

$= {( \sqrt{15} )}^{3} + 3 {(4)}^{2} \sqrt{15} + {( \sqrt{15} )}^{3} + 3 {(4)}^{2} \sqrt{15}$

$= 15 \sqrt{15} + 48 \sqrt{15} + 15 \sqrt{15} + 48 \sqrt{15}$

$= 126 \sqrt{15}$