1. **Stating the problem:** Given the equation $$x^4 + 1 = \frac{5}{8} x^2$$, find the values of: - $$x + \frac{1}{x}$$ - $$(x - \frac{1}{x})^3$$ - $$x^2 + \frac{1}{x^5}$$ 2. **Rewrite the given equation:** Move all terms to one side: $$x^4 - \frac{5}{8} x^2 + 1 = 0$$ Let $$y = x^2$$ to simplify the quartic to a quadratic: $$y^2 - \frac{5}{8} y + 1 = 0$$ 3. **Solve for $$y$$:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-\frac{5}{8}$$, $$c=1$$: $$b^2 = \left(-\frac{5}{8}\right)^2 = \frac{25}{64}$$ $$4ac = 4 \times 1 \times 1 = 4$$ Discriminant: $$\Delta = \frac{25}{64} - 4 = \frac{25}{64} - \frac{256}{64} = -\frac{231}{64} < 0$$ Since the discriminant is negative, $$x^2$$ has no real roots, implying $$x$$ is complex. 4. **Find $$x + \frac{1}{x}$$:** Multiply both sides of the original equation by $$\frac{1}{x^2}$$ (assuming $$x \neq 0$$): $$x^2 + \frac{1}{x^2} = \frac{5}{8}$$ Note that: $$(x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$ Thus: $$(x + \frac{1}{x})^2 = \frac{5}{8} + 2 = \frac{5}{8} + \frac{16}{8} = \frac{21}{8}$$ So: $$x + \frac{1}{x} = \pm \sqrt{\frac{21}{8}} = \pm \frac{\sqrt{21}}{2 \sqrt{2}} = \pm \frac{\sqrt{42}}{4}$$ 5. **Find $$(x - \frac{1}{x})^3$$:** First find $$x - \frac{1}{x}$$: $$(x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2} = \frac{5}{8} - 2 = \frac{5}{8} - \frac{16}{8} = -\frac{11}{8}$$ Since this is negative, $$x - \frac{1}{x}$$ is imaginary. Let: $$u = x - \frac{1}{x}$$ Then: $$u^2 = -\frac{11}{8}\implies u = \pm i \sqrt{\frac{11}{8}} = \pm i \frac{\sqrt{11}}{2 \sqrt{2}} = \pm i \frac{\sqrt{22}}{4}$$ Cube this: $$(x - \frac{1}{x})^3 = u^3 = (\pm i \frac{\sqrt{22}}{4})^3 = \pm i^3 \left( \frac{\sqrt{22}}{4} \right)^3 = \pm (-i) \frac{(\sqrt{22})^3}{64} = \mp i \frac{22 \sqrt{22}}{64} = \mp i \frac{11 \sqrt{22}}{32}$$ 6. **Find $$x^2 + \frac{1}{x^5}$$:** Write $$\frac{1}{x^5} = \frac{1}{x^2} \cdot \frac{1}{x^3}$$. We know: $$x^2 = y$$ We have from step 4: $$x^2 + \frac{1}{x^2} = \frac{5}{8}$$ We want: $$x^2 + \frac{1}{x^5} = x^2 + \frac{1}{x^2} \cdot \frac{1}{x^3}$$ We don't know $$\frac{1}{x^3}$$ directly. Instead, from the initial equation, recall: $$x^4 + 1 = \frac{5}{8} x^2$$ Divide both sides by $$x^5$$: $$\frac{x^4}{x^5} + \frac{1}{x^5} = \frac{5}{8} \frac{x^2}{x^5}$$ Simplify: $$\frac{1}{x} + \frac{1}{x^5} = \frac{5}{8} \frac{1}{x^3}$$ Thus: $$\frac{1}{x^5} = \frac{5}{8} \frac{1}{x^3} - \frac{1}{x}$$ Rewrite $$x^2 + \frac{1}{x^5}$$: $$x^2 + \frac{1}{x^5} = x^2 + \frac{5}{8} \frac{1}{x^3} - \frac{1}{x}$$ This expression depends on higher powers of $$1/x$$ and is not directly computable without knowing $$x$$ explicitly. Given the complexity and initial imaginary results, the expression cannot be simplified further without additional information about $$x$$. **Final answers:** - $$x + \frac{1}{x} = \pm \frac{\sqrt{42}}{4}$$ - $$(x - \frac{1}{x})^3 = \mp i \frac{11 \sqrt{22}}{32}$$ (imaginary) - $$x^2 + \frac{1}{x^5}$$ cannot be simplified further with given data and is generally complex.