1. **Problem Statement:** Solve the equation $$10(x^4+1) + 63(x^3 - x) + 52x^2 = 0$$ 2. **Step 1: Rearrange and divide by $x^2$ (assuming $x \neq 0$):** $$10(x^2 + \frac{1}{x^2}) + 63(x - \frac{1}{x}) + 52 = 0$$ 3. **Step 2: Use the identity:** $$x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2$$ 4. **Step 3: Substitute $y = x - \frac{1}{x}$:** $$10(y^2 + 2) + 63y + 52 = 0$$ 5. **Step 4: Simplify:** $$10y^2 + 20 + 63y + 52 = 0 \Rightarrow 10y^2 + 63y + 72 = 0$$ 6. **Step 5: Factor the quadratic:** $$10y^2 + 63y + 72 = (5y + 24)(2y + 3) = 0$$ 7. **Step 6: Solve for $y$:** $$y = -\frac{24}{5} \quad \text{or} \quad y = -\frac{3}{2}$$ 8. **Step 7: Solve for $x$ when $y = x - \frac{1}{x}$:** (i) For $y = -\frac{24}{5}$: $$x - \frac{1}{x} = -\frac{24}{5} \Rightarrow 5x^2 + 24x - 5 = 0$$ Solve quadratic: $$x = \frac{-24 \pm \sqrt{24^2 - 4 \cdot 5 \cdot (-5)}}{2 \cdot 5} = \frac{-24 \pm \sqrt{576 + 100}}{10} = \frac{-24 \pm 26}{10}$$ Solutions: $$x = -5 \quad \text{or} \quad x = \frac{1}{5}$$ (ii) For $y = -\frac{3}{2}$: $$x - \frac{1}{x} = -\frac{3}{2} \Rightarrow 2x^2 + 3x - 2 = 0$$ Solve quadratic: $$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm 5}{4}$$ Solutions: $$x = -2 \quad \text{or} \quad x = \frac{1}{2}$$ **Final answer:** $$x = -5, \frac{1}{5}, -2, \frac{1}{2}$$