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

$a=\frac{1}{3-2\sqrt{2}}$
a=$=\frac{3+2\sqrt{2}}{9-8}$ $(Multiplyingby3+2\sqrt{2}$)
a=$3+2\sqrt{2}$
$a^2=9+8+12\sqrt{2}$
$b=\frac{1}{3+2\sqrt{2}}$
$b=\frac{3-2\sqrt{2}}{9-8}$ $(Multiplyingby3-2\sqrt{2}$)
b=$3-2\sqrt{2}$
$b^2=9+8-12\sqrt{2}$
$a^3+b^3$
$=(a+b)(a^2+b^2-ab)$
$=6(9+8+12\sqrt{2}+9+8-12\sqrt{2}-1)$
$=6(34-1)$
$=6\times 33=198$