If x = 3 - 1 3 and y = 3 + 1 3 then the value of x^2y + y^2x is?

Answer: Option B

$$\begin{array}{rl}& x=\sqrt{3}-\frac{1}{\sqrt{3}}\text{ and }y=\sqrt{3}+\frac{1}{\sqrt{3}}\\ & \Rightarrow \frac{x^2}{y}+\frac{y^2}{x}\\ & =\frac{x^3+y^3}{xy}\\ & =\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy}\\ & \therefore x+y\\ & =\sqrt{3}-\frac{1}{\sqrt{3}}+\sqrt{3}+\frac{1}{\sqrt{3}}\\ & =2\sqrt{3}\\ & \therefore xy\\ & =\sqrt{3}-\frac{1}{\sqrt{3}}\times \sqrt{3}+\frac{1}{\sqrt{3}}\\ & =3-\frac{1}{3}\\ & =\frac{8}{3}\\ & \Rightarrow \frac{\left(x+y\right)\left(x^2+y^2+2xy-2xy-xy\right)}{xy}\\ & \Rightarrow \frac{\left(x+y\right)\left({\left(x+y\right)}^2-3xy\right)}{xy}\\ & \Rightarrow \frac{2\sqrt{3}\left({\left(2\sqrt{3}\right)}^2-3\times \frac{8}{3}\right)}{\frac{8}{3}}\\ & \Rightarrow \frac{2\sqrt{3}\left(12-8\right)}{\frac{8}{3}}\\ & \Rightarrow \frac{2\times 3\sqrt{3}\left(4\right)}{8}\\ & \Rightarrow 3\sqrt{3}\end{array}$$