Question

State True or False.

If $x=3-2\sqrt{2}$, then the value of $x^3-\frac{1}{x^3}$$=-140\sqrt{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

$x=3-2\sqrt{2}$
$\frac{1}{x}=\frac{1}{3-2\sqrt{2}}$
$\frac{1}{x}=\frac{3+2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})}$
$=\frac{3+2\sqrt{2}}{3^2-(2\sqrt{2}{)}^2}$
$=\frac{3+2\sqrt{2}}{9-8}$
$=3+2\sqrt{2}$
Now,
$x-\frac{1}{x}=3-2\sqrt{2}-(3+2\sqrt{2})$
$=-4\sqrt{2}$
We know,
$x^2-\frac{1}{x^2}=(x+\frac{1}{x}{)}^2-2(x\left)\right(\frac{1}{x})$
$=(3-2\sqrt{2}+3+2\sqrt{2}{)}^2-2$
$=6^2-2$
$=36-2$
$=34$
Now,
$x^3-\frac{1}{x^3}=(x-\frac{1}{x})[x^2+\frac{1}{x^2}+(x\left)\right(\frac{1}{x}\left)\right]$
$=(-4\sqrt{2})(34+1)$
$=(-4\sqrt{2})35$
$=-140\sqrt{2}$