Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Option 1: $\sqrt{3}$
Option 2: $3\sqrt{3}$
Option 3: $16\sqrt{3}$
Option 4: $2\sqrt{3}$
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Correct Answer: $3\sqrt{3}$
Solution : Given: $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$ $⇒x=\frac{2}{\sqrt{3}}, y =\frac{4}{\sqrt{3}}$ Now, $\frac{x^2}{y}+\frac{y^2}{x}$ $=\frac{(\frac{2}{\sqrt{3}})^2}{\frac{4}{\sqrt{3}}}+\frac{(\frac{4}{\sqrt{3}})^2}{\frac{2}{\sqrt{3}}}$ $=\frac{1}{\sqrt{3}}+\frac{8}{\sqrt{3}}$ $=3\sqrt{3}$ Hence, the correct answer is $3\sqrt{3}$.
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