A) \[\sqrt{3}\]
B) \[3\sqrt{3}\]
C) \[16\sqrt{3}\]
D) \[2\sqrt{3}\]
Correct Answer: B
Solution :
(b): \[x=\sqrt{3}+\frac{1}{\sqrt{3}}\,and\,y=\sqrt{3}-\frac{1}{\sqrt{3}}\] \[\therefore x+y=\sqrt{3}+\frac{1}{\sqrt{3}}+\sqrt{3}-\frac{1}{\sqrt{3}}=2\sqrt{3}\] And \[xy-\left( \sqrt{3}+\frac{1}{\sqrt{3}} \right)\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)=\frac{8}{3}\] \[\therefore \frac{{{x}^{2}}}{y}+\frac{{{y}^{2}}}{x}=\frac{{{x}^{3}}+{{y}^{3}}}{xy}=\frac{{{\left( x+y \right)}^{3}}-3xy\left( x+y \right)}{xy}\] \[=\frac{24\sqrt{3}-16\sqrt{3}}{\frac{8}{3}}=\frac{8\left( 3\sqrt{3}-2\sqrt{3} \right)}{\frac{8}{3}}=9\sqrt{3}-6\sqrt{3}=3\sqrt{3}\]
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