Question

if x=1/4-√15,y=1/4+√15,then the value of x³+y³ is

Answer 1

Given-

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

And,

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

To find -

The value of x³ + y³

Solution -

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

By rationalising the denominator we will -

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

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

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

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

$x = 4 + \sqrt{15}$

And,

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

By rationalising the denominator we will get -

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

$y = \frac{4 - \sqrt{15} }{ {4}^{2} - { \sqrt{15} }^{2} }$

$y = \frac{4 - \sqrt{15} }{16 - 15}$

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

$y = 4 - \sqrt{15}$

Now,

=x³+y³

Putting the values we will get -

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

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

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

$= 64 + 180 + 64 + 180$

$= 128 + 360$

$= 488$