If $$x$$ = $$\sqrt 3 - \frac{1}{{\sqrt 3 }}$$ and $$y$$ = $$\sqrt 3 + \frac{1}{{\sqrt 3 }}$$ then the value of $$\frac{{{x^2}}}{y} + \frac{{{y^2}}}{x}$$ is? Options: A . &nbsp$$\sqrt 3 $$ B . &nbsp$${\text{3}}\sqrt 3 $$ C . &nbsp$${\text{16}}\sqrt 3 $$ D . &nbsp$${\text{2}}\sqrt 3 $$

If $$x$$ = $$\sqrt 3 - \frac{1}{{\sqrt 3 }}$$ and $$y$$ = $$\sqrt 3 + \frac{1

Question

If $$x$$ = $$\sqrt 3 - \frac{1}{{\sqrt 3 }}$$ and $$y$$ = $$\sqrt 3 + \frac{1}{{\sqrt 3 }}$$ then the value of $$\frac{{{x^2}}}{y} + \frac{{{y^2}}}{x}$$ is?

Options:

A . $$\sqrt 3 $$

B . $${\text{3}}\sqrt 3 $$

C . $${\text{16}}\sqrt 3 $$

D . $${\text{2}}\sqrt 3 $$

Answer: Option B
$$\eqalign{
& x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr
& = \frac{{{x^3} + {y^3}}}{{xy}} \cr
& = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr
& \therefore x + y \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 2\sqrt 3 \cr
& \therefore xy \cr
& = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr
& = 3 - \frac{1}{3} \cr
& = \frac{8}{3} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr
& \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr
& \Rightarrow 3\sqrt 3 \cr} $$

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