A) 189
B) 180
C) 108
D) 198
Correct Answer: D
Solution :
We know that, \[{{a}^{3}}+{{b}^{3}}={{(a+b)}^{3}}-3ab\,\,(a+b)\] Now, \[{{(3+2\sqrt{2})}^{-\,\,3}}+\,\,{{(3-2\sqrt{2})}^{-\,\,3}}\] \[=\frac{1}{{{(3+2\sqrt{2})}^{3}}}+\frac{1}{{{(3-2\sqrt{2})}^{3}}}=\frac{{{(3-2\sqrt{2})}^{3}}+{{(3+2\sqrt{2})}^{3}}}{{{(3+2\sqrt{2})}^{3}}\times {{(3-2\sqrt{2})}^{3}}}\]\[=\frac{\begin{align} & {{(3-2\sqrt{2}+3+2\sqrt{2})}^{3}}-3\,\,(3-2\sqrt{2}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3+2\sqrt{2})(3-2\sqrt{2}+3+2\sqrt{2}) \\ \end{align}}{{{[(3+2\sqrt{2})(3-2\sqrt{2})]}^{3}}}\]
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