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