Aptitude Question ID: 107124

(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.

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