A) \[10\sqrt{2}\]
B) \[14\sqrt{2}\]
C) \[22\sqrt{2}\]
D) \[8\sqrt{2}\]
Correct Answer: C
Solution :
\[x=\sqrt{3}+\sqrt{2}\] \[\therefore \] \[\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\] \[=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}\] \[\therefore \]\[x-\frac{1}{x}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\] \[\therefore \] \[{{x}^{3}}-\frac{1}{{{x}^{3}}}={{\left( x-\frac{1}{x} \right)}^{3}}+3\left( x-\frac{1}{x} \right)\] \[={{(2\sqrt{2})}^{3}}+3\times 2\sqrt{2}\] \[=16\sqrt{2}+6\sqrt{2}=22\sqrt{2}\]
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