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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