1. Stating the problem: Solve the equation $X^{39} + 6 = X^{50} - 6$ for $X$. 2. Rearrange the equation to isolate terms: $$X^{39} + 6 = X^{50} - 6 \implies X^{39} - X^{50} = -6 - 6$$ $$X^{39} - X^{50} = -12$$ 3. Factor out the common term $X^{39}$: $$X^{39} - X^{50} = X^{39}(1 - X^{11}) = -12$$ 4. Rewrite the equation: $$X^{39}(1 - X^{11}) = -12$$ 5. This is a nonlinear equation in $X$. To find real solutions, consider possible values of $X$. 6. Check if $X=1$ is a solution: $$1^{39}(1 - 1^{11}) = 1 \times 0 = 0 \neq -12$$ 7. Check if $X=-1$ is a solution: $$(-1)^{39}(1 - (-1)^{11}) = (-1)(1 - (-1)) = (-1)(1 + 1) = (-1)(2) = -2 \neq -12$$ 8. Since the equation is complex, rewrite as: $$X^{39}(1 - X^{11}) + 12 = 0$$ 9. Let $Y = X^{11}$, then $X^{39} = (X^{11})^{3} = Y^{3}$, so: $$Y^{3}(1 - Y) + 12 = 0$$ $$Y^{3} - Y^{4} + 12 = 0$$ 10. Rearranged: $$-Y^{4} + Y^{3} + 12 = 0$$ 11. Multiply both sides by $-1$: $$Y^{4} - Y^{3} - 12 = 0$$ 12. Solve quartic equation: $$Y^{4} - Y^{3} - 12 = 0$$ 13. Try rational roots using factors of 12: $\\pm1, \\pm2, \\pm3, \\pm4, \\pm6, \\pm12$ 14. Test $Y=2$: $$2^{4} - 2^{3} - 12 = 16 - 8 - 12 = -4 \neq 0$$ 15. Test $Y=3$: $$3^{4} - 3^{3} - 12 = 81 - 27 - 12 = 42 \neq 0$$ 16. Test $Y=4$: $$4^{4} - 4^{3} - 12 = 256 - 64 - 12 = 180 \neq 0$$ 17. Test $Y=-1$: $$(-1)^{4} - (-1)^{3} - 12 = 1 + 1 - 12 = -10 \neq 0$$ 18. Test $Y=-2$: $$(-2)^{4} - (-2)^{3} - 12 = 16 + 8 - 12 = 12 \neq 0$$ 19. No simple rational roots found; numerical methods or graphing needed for exact roots. 20. Once $Y$ is found, solve for $X$: $$X^{11} = Y \implies X = \sqrt[11]{Y}$$ Final answer: The equation reduces to solving the quartic $$Y^{4} - Y^{3} - 12 = 0$$ for $Y = X^{11}$, then finding $X = \sqrt[11]{Y}$. Exact roots require numerical methods.