Question

please answer this question. See pic

Answer 1

$Given : x = 2 + \sqrt{3}$

$= \ \textgreater \ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } * \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$= \ \textgreater \ \frac{1}{x} = \frac{2 - \sqrt{3} }{2^2 - ( \sqrt{3} )^2}$

$= \ \textgreater \ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}$

$= \ \textgreater \ \frac{1}{x} = 2 - \sqrt{3}$


Now,

$= \ \textgreater \ x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3}$

= > 4.


On cubing both sides, we get

= > (x + 1/x)^3 = (4)^3

= > x^3 + 1/x^3 + 3 * x^3 * 1/x^3(x + 1/x) = 64

= > x^3 + 1/x^3 + 3(4) = 64

= > x^3 + 1/x^3 + 12 = 64

= > x^3 + 1/x^3 = 64 - 12

= > x^3 + 1/x^3 = 52.


Hope this helps!

Answer 2

Hi,

Please see the attached file !


Thanks