Question

If x= 2 +√3, find the value of x³ +$\frac{1}{x³}$. Please give method and answer.

Answer 1

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

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

$= \ \textgreater \ \frac{2 - \sqrt{3} }{(2 + \sqrt{3})(2 - \sqrt{3}) }$

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

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

$= \ \textgreater \ 2 - \sqrt{3}$


Now,

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

= > (x + 1/x) = 4

Now,

On cubing both sides, we get

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

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