Question

If x=1/4-√15 , y=1/4+√15 then x cube + y cube =

Answer 1

Given,
$x = \frac{1}{4 - \sqrt{15} }$
and,
$y = \frac{1}{4 + \sqrt{15} }$
We have to find,
${x}^{3} + {y}^{3}$

Finding :

$x = \frac{1}{4 - \sqrt{15} }$
Rationalising the denominator,

$= > x = \frac{1}{4 - \sqrt{15} } \times \frac{4 + \sqrt{15} }{4 + \sqrt{15} }$
$= > x = \frac{1 \times 4 + \sqrt{15} }{ ({4})^{2} - {( \sqrt{15}) }^{2} }$
$= > x = \frac{4 + \sqrt{15} }{16 - 15}$
$= > x = \frac{4 + \sqrt{15} }{1}$
$= > x = 4 + \sqrt{15} \: \: \:$
Therefore, x = 4 + √15 (I)


Again,

$y = \frac{1}{4 + \sqrt{15} }$
Rationalising the denominator,

$= > y = \frac{1}{4 + \sqrt{15} } \times \frac{4 - \sqrt{15} }{4 - \sqrt{15} }$
$= > y = \frac{1 \times 4 - \sqrt{15} }{ ({4})^{2} - { (\sqrt{15} )}^{2} }$
$= > y = \frac{4 - \sqrt{15} }{16 - 15}$
$= > y = \frac{4 - \sqrt{15} }{1}$
$= > y = 4 - \sqrt{15}$
Therefore, y = 4 - √15. (II)

Now,
x + y
= (4 + √15) + (4 - √15)
= 4 + √15 + 4 - √15
= 4 + 4 + 0
= 8

Further,
xy
= (4 + √15) (4 - √15)
= 16 + 4√15 - 4√15 - 15
= 16 - 15
= 1


Now, we know that,

${x}^{3} + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$
$= (8)({4 + \sqrt{15}})^{2}+ {(4 - \sqrt{15} })^{2} - 1$
$= (8)(16 + 15) + (16 - 15) - (1)$
$= (8)(31) + (1) - (1)$
$= 8 \times 31$
$= 248$

That's your answer.

Hope it'll help.. :-D

Answer 2

Given,

and,

We have to find,

Finding :

Rationalising the denominator,

Therefore, x = 4 + √15 (I)

Again,

Rationalising the denominator,

Therefore, y = 4 - √15. (II)

Now,
x + y
= (4 + √15) + (4 - √15)
= 4 + √15 + 4 - √15
= 4 + 4 + 0
= 8

Further,
xy
= (4 + √15) (4 - √15)
= 16 + 4√15 - 4√15 - 15
= 16 - 15
= 1

Now, we know that,

248....

That's your answer.