Question

If $a=\frac{1}{3-2\sqrt{2}},b=\frac{1}{3+2\sqrt{2}}$, then the value of $a^3+b^3$ is:

• A. $194$
• B. $196$
• C. $198$
• D. $200$

Answer 1

The correct option is C $198$

$a=\frac{1}{3-2\sqrt{2}},b=\frac{1}{3+2\sqrt{2}}$,

$a=\frac{3+2\sqrt{2}}{(3-2\sqrt{2})(3+2\sqrt{2})},b=\frac{3-2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}$

$a=3+2\sqrt{2},b=3-2\sqrt{2}$

Now, $a^3+b^3=(a+b{)}^3-3ab(a+b)$
$=(3+2\sqrt{2}+3-2\sqrt{2}{)}^3-3(3+2\sqrt{2}\left)\right(3-2\sqrt{2}\left)\right(3+2\sqrt{2}+3-2\sqrt{2})$
$=6^3-3(9-8)\left(6\right)$
$=216-18$
$=198$