Question: Given $x + y = \sqrt{3}$, $x^2 + y^2 = 5$, evaluate the following expressions. (1) $xy$ (2) $x^3 + y^3$ (3) $\frac{y}{x} + \frac{x}{y}$ (4) $\frac{y}{x^2} + \frac{x}{y^2}$

1. **State the problem:** Given $x + y = \sqrt{3}$ and $x^2 + y^2 = 5$, find: (1) $xy$ (2) $x^3 + y^3$ (3) $\frac{y}{x} + \frac{x}{y}$ (4) $\frac{y}{x^2} + \frac{x}{y^2}$ --- 2. **Find $xy$ using the identity:** $$ (x + y)^2 = x^2 + 2xy + y^2 $$ Substitute the known values: $$ (\sqrt{3})^2 = 5 + 2xy $$ $$ 3 = 5 + 2xy $$ Solve for $xy$: $$ 2xy = 3 - 5 = -2 $$ $$ xy = \frac{-2}{2} = -1 $$ --- 3. **Find $x^3 + y^3$ using the formula:** $$ x^3 + y^3 = (x + y)(x^2 - xy + y^2) $$ Calculate $x^2 - xy + y^2$: $$ x^2 - xy + y^2 = 5 - (-1) = 5 + 1 = 6 $$ Then: $$ x^3 + y^3 = (\sqrt{3})(6) = 6\sqrt{3} $$ --- 4. **Find $\frac{y}{x} + \frac{x}{y}$:** Rewrite as: $$ \frac{y}{x} + \frac{x}{y} = \frac{y^2 + x^2}{xy} $$ Substitute known values: $$ = \frac{5}{-1} = -5 $$ --- 5. **Find $\frac{y}{x^2} + \frac{x}{y^2}$:** Rewrite as: $$ \frac{y}{x^2} + \frac{x}{y^2} = \frac{y^3 + x^3}{x^2 y^2} $$ Calculate denominator: $$ x^2 y^2 = (xy)^2 = (-1)^2 = 1 $$ Substitute numerator and denominator: $$ = \frac{6\sqrt{3}}{1} = 6\sqrt{3} $$ --- **Final answers:** 1. $xy = -1$ 2. $x^3 + y^3 = 6\sqrt{3}$ 3. $\frac{y}{x} + \frac{x}{y} = -5$ 4. $\frac{y}{x^2} + \frac{x}{y^2} = 6\sqrt{3}$