Question

If $x=3-2\sqrt{2}$ then $(x+\frac{1}{x}{)}^3$ = __________________

Answer 1

$x=3-2\sqrt{2}$ ...(i)
$\Rightarrow \frac{1}{x}=\frac{1}{3-2\sqrt{2}}$
Rationalizing factor of $3-2\sqrt{2}$ is $3+2\sqrt{2}$

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

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

$=\frac{3+2\sqrt{2}}{9-8}$

$=3+2\sqrt{2}$ ...(ii)

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

$=6$

$\Rightarrow (x+\frac{1}{x}{)}^3=(6{)}^3$

$=216$