$(3+2\sqrt{2})^{-3}+(3-2\sqrt{2})^{-3}$ is equal to :
[A]108
[B]180
[C]189
[D]198

198
Becaues $a^{3}+b^{3} = (a+b)^{3}-3abc(a+b)$
$(3+2\sqrt{2})^{-3}+(3-2\sqrt{2})^{-3}$
$= \frac{1}{(3+2\sqrt{2})^{3}}+\frac{1}{(3-2\sqrt{2})^{3}}$
$= \frac{(3-2\sqrt{2})^{3}+(3+2\sqrt{2})^{3}}{(3+2\sqrt{2})^{3}\times (3-2\sqrt{2})^{3}}$
$= \frac{(3-2\sqrt{2}+3+2\sqrt{2})^{3}-3(3-2\sqrt{2})(3+2\sqrt{2})(3-2\sqrt{2}+3+2\sqrt{2})}{\left \{ (3+2\sqrt{2})(3-2\sqrt{2}) \right \}^{3}}$
$= \frac{(6)^{3}-3(9-8)(6)}{1} = 216-18 = 198$
Hence option [D] is correct answer.