1. **Problem statement:** Solve for $x$ in the equation $$x - \frac{1}{x} = \sqrt{x} + \frac{1}{\sqrt{x}}$$ and then evaluate $$x^y + x^{-y}$$. 2. **Step 1: Simplify the given equation.** Let $t = \sqrt{x}$, so $x = t^2$ and $\frac{1}{x} = \frac{1}{t^2}$. 3. Substitute into the equation: $$t^2 - \frac{1}{t^2} = t + \frac{1}{t}$$ 4. Multiply both sides by $t^2$ to clear denominators: $$t^4 - 1 = t^3 + t$$ 5. Rearrange all terms to one side: $$t^4 - t^3 - t - 1 = 0$$ 6. Factor by grouping: $$t^3(t - 1) - 1(t + 1) = 0$$ 7. This does not factor nicely, so try rational roots. Test $t=1$: $$1 - 1 - 1 - 1 = -2 \neq 0$$ Test $t=-1$: $$1 + 1 + 1 - 1 = 2 \neq 0$$ 8. Use substitution $u = t - 1$ or numerical methods to approximate roots, but since $t = \sqrt{x} \geq 0$, check positive roots. 9. Alternatively, rewrite original equation as: $$x - \frac{1}{x} - \sqrt{x} - \frac{1}{\sqrt{x}} = 0$$ 10. Multiply both sides by $\sqrt{x}$: $$x\sqrt{x} - \frac{\sqrt{x}}{x} - x - 1 = 0$$ Since $x\sqrt{x} = x^{3/2}$ and $\frac{\sqrt{x}}{x} = x^{-1/2}$, the equation becomes: $$x^{3/2} - x^{-1/2} - x - 1 = 0$$ 11. This is complicated; however, the problem likely expects to express $x^y + x^{-y}$ in terms of $\sqrt{x} + \frac{1}{\sqrt{x}}$. 12. Note that: $$x^y + x^{-y} = (\sqrt{x})^{2y} + (\sqrt{x})^{-2y}$$ 13. Let $a = \sqrt{x} + \frac{1}{\sqrt{x}}$, then by the identity for powers: $$x^y + x^{-y} = (\sqrt{x})^{2y} + (\sqrt{x})^{-2y} = 2T_{2y}(\frac{a}{2})$$ where $T_n$ is the Chebyshev polynomial of the first kind. 14. Since the problem does not provide $y$ or further conditions, the expression for $x^y + x^{-y}$ is: $$x^y + x^{-y} = 2T_{2y}\left(\frac{\sqrt{x} + \frac{1}{\sqrt{x}}}{2}\right)$$ **Final answer:** $$x^y + x^{-y} = 2T_{2y}\left(\frac{\sqrt{x} + \frac{1}{\sqrt{x}}}{2}\right)$$ This expresses the quantity in terms of $\sqrt{x} + \frac{1}{\sqrt{x}}$ as given in the problem.