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1. The value of ${(3 + 2 \sqrt{2})}^{- 3}$ $+ {(3 - 2 \sqrt{2})}^{- 3} = ?$

A. 189

B. 180

C. 108

D. 198

Solution:

\begin{aligned} {(3 + 2 \sqrt{2})}^{- 3} + {(3 - 2 \sqrt{2})}^{- 3} \\ = {(\frac{1}{3 + 2 \sqrt{2}})}^{3} + {(\frac{1}{3 - 2 \sqrt{2}})}^{3} \end{aligned}

= {(\frac{1}{(3 - 2 \sqrt{2})} \times \frac{3 + 2 \sqrt{2}}{3 + 2 \sqrt{2}})}^{3} +

{(\frac{1}{(3 + 2 \sqrt{2})} \times \frac{3 - 2 \sqrt{2}}{3 - 2 \sqrt{2}})}^{3}

\begin{aligned} = {(\frac{3 - 2 \sqrt{2}}{9 - 8})}^{3} + {(\frac{3 + 2 \sqrt{2}}{9 - 8})}^{3} \\ = {(3 - 2 \sqrt{2})}^{3} + {(3 + 2 \sqrt{2})}^{3} \\ a = 3 - 2 \sqrt{2} \\ b = 3 + 2 \sqrt{2} \\ [∵ a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)] \\ = (3 - 2 \sqrt{2} + 3 + 2 \sqrt{2}) (17 + 17 - 1) \\ = (6) (33) \\ = 198 \end{aligned}

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