A) 0
B) \[-\sqrt{2}\]
C) \[2\sqrt{2}\]
D) \[3\sqrt{2}\]
Correct Answer: C
Solution :
(c): \[x={{\left( \sqrt{2}+1 \right)}^{\frac{-1}{3}}}\,\,\,\,\Rightarrow \frac{1}{{{x}^{3}}}=\sqrt{2}+1\] And \[{{x}^{3}}=\frac{1}{\sqrt{2}+1}=\frac{1\left( \sqrt{2}-1 \right)}{\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)}=\sqrt{2}-1\] \[\therefore {{x}^{3}}+\frac{1}{{{x}^{3}}}\,\,\,\,=\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\]
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