Question

pls answer question .I will mark him as brainliest

Answer 1

your solution friend hope it helps.

Answer 2

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

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

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

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

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

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

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

Now,

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

= > -4.

Now,

$= \ \textgreater \ x + (1/x) = (-4)^3$

= > -64


Hope this helps!