A) 2
B) 4
C) 8
D) 9
Correct Answer: B
Solution :
(b): Given \[x=\sqrt[3]{2+\sqrt{3}}\] Take cube both side, are get \[{{x}^{3}}=2+\sqrt{3}\] ________ (1) \[\frac{1}{{{x}^{3}}}=\frac{1}{\left( 2+\sqrt{3} \right)}\times \frac{\left( 2-\sqrt{3} \right)}{\left( 2-\sqrt{3} \right)}\] ______(2) \[\frac{1}{{{x}^{3}}}=2-\sqrt{3}\] So, \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=2+\sqrt{3}+2-\sqrt{3}=4\]You need to login to perform this action.
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