1. **State the problem:** Solve the rational equation $$\frac{1}{5 - x^4} + \frac{1}{x^4 - 3} = 2$$. 2. **Rewrite the equation:** Notice that the denominators are $$5 - x^4$$ and $$x^4 - 3$$. 3. **Find a common denominator:** The common denominator is $$(5 - x^4)(x^4 - 3)$$. 4. **Rewrite each fraction with the common denominator:** $$\frac{1}{5 - x^4} = \frac{x^4 - 3}{(5 - x^4)(x^4 - 3)}$$ $$\frac{1}{x^4 - 3} = \frac{5 - x^4}{(5 - x^4)(x^4 - 3)}$$ 5. **Add the fractions:** $$\frac{x^4 - 3}{(5 - x^4)(x^4 - 3)} + \frac{5 - x^4}{(5 - x^4)(x^4 - 3)} = \frac{(x^4 - 3) + (5 - x^4)}{(5 - x^4)(x^4 - 3)} = \frac{2}{(5 - x^4)(x^4 - 3)}$$ 6. **Set the equation equal to 2:** $$\frac{2}{(5 - x^4)(x^4 - 3)} = 2$$ 7. **Multiply both sides by the denominator to clear fractions:** $$2 = 2(5 - x^4)(x^4 - 3)$$ 8. **Divide both sides by 2:** $$\cancel{2} = \cancel{2}(5 - x^4)(x^4 - 3) \implies 1 = (5 - x^4)(x^4 - 3)$$ 9. **Expand the right side:** $$1 = 5x^4 - 15 - x^8 + 3x^4 = -x^8 + 8x^4 - 15$$ 10. **Bring all terms to one side:** $$-x^8 + 8x^4 - 15 - 1 = 0 \implies -x^8 + 8x^4 - 16 = 0$$ 11. **Multiply both sides by -1 to simplify:** $$x^8 - 8x^4 + 16 = 0$$ 12. **Substitute $y = x^4$ to reduce degree:** $$y^2 - 8y + 16 = 0$$ 13. **Solve quadratic equation:** $$y = \frac{8 \pm \sqrt{64 - 64}}{2} = \frac{8 \pm 0}{2} = 4$$ 14. **Back-substitute:** $$x^4 = 4$$ 15. **Solve for $x$:** $$x = \pm \sqrt{2}$$ 16. **Check for restrictions:** Denominators cannot be zero: - $5 - x^4 \neq 0 \implies x^4 \neq 5$ - $x^4 - 3 \neq 0 \implies x^4 \neq 3$ Since $x^4 = 4$ is allowed, both $x = \pm \sqrt{2}$ are valid solutions. **Final answer:** $$x = \pm \sqrt{2}$$