1. **State the problem:** Solve the rational equation $$\frac{12}{x^2 - 2x} + 3 = \frac{6}{x - 2}$$. 2. **Identify the domain restrictions:** The denominators cannot be zero. - For $$x^2 - 2x = x(x-2)$$, so $$x \neq 0$$ and $$x \neq 2$$. 3. **Rewrite the equation:** $$\frac{12}{x(x-2)} + 3 = \frac{6}{x-2}$$ 4. **Express 3 as a fraction with denominator $$x(x-2)$$ to combine terms:** $$3 = \frac{3x(x-2)}{x(x-2)}$$ 5. **Rewrite the equation with common denominators:** $$\frac{12}{x(x-2)} + \frac{3x(x-2)}{x(x-2)} = \frac{6}{x-2}$$ 6. **Combine the left side:** $$\frac{12 + 3x(x-2)}{x(x-2)} = \frac{6}{x-2}$$ 7. **Cross-multiply to clear denominators:** $$\left(12 + 3x(x-2)\right)(x-2) = 6 \cdot x(x-2)$$ 8. **Simplify the left side numerator:** $$3x(x-2) = 3x^2 - 6x$$ So numerator is $$12 + 3x^2 - 6x = 3x^2 - 6x + 12$$ 9. **Multiply numerator by $$x-2$$:** $$\left(3x^2 - 6x + 12\right)(x-2) = 6x(x-2)$$ 10. **Expand left side:** $$3x^2 \cdot x = 3x^3$$ $$3x^2 \cdot (-2) = -6x^2$$ $$-6x \cdot x = -6x^2$$ $$-6x \cdot (-2) = 12x$$ $$12 \cdot x = 12x$$ $$12 \cdot (-2) = -24$$ Sum: $$3x^3 - 6x^2 - 6x^2 + 12x + 12x - 24 = 3x^3 - 12x^2 + 24x - 24$$ 11. **Expand right side:** $$6x(x-2) = 6x^2 - 12x$$ 12. **Set equation:** $$3x^3 - 12x^2 + 24x - 24 = 6x^2 - 12x$$ 13. **Bring all terms to one side:** $$3x^3 - 12x^2 + 24x - 24 - 6x^2 + 12x = 0$$ Simplify: $$3x^3 - 18x^2 + 36x - 24 = 0$$ 14. **Divide entire equation by 3:** $$\cancel{3}x^3 - \cancel{3}6x^2 + \cancel{3}12x - \cancel{3}8 = 0$$ $$x^3 - 6x^2 + 12x - 8 = 0$$ 15. **Factor the cubic:** Try to find roots by Rational Root Theorem. Try $$x=2$$: $$2^3 - 6(2)^2 + 12(2) - 8 = 8 - 24 + 24 - 8 = 0$$ So $$x=2$$ is a root. 16. **Divide polynomial by $$x-2$$:** Using synthetic division: Coefficients: 1 | -6 | 12 | -8 Bring down 1. Multiply 1*2=2, add to -6 = -4. Multiply -4*2 = -8, add to 12 = 4. Multiply 4*2=8, add to -8=0. Quotient: $$x^2 - 4x + 4$$ 17. **Factor quotient:** $$x^2 - 4x + 4 = (x-2)^2$$ 18. **Full factorization:** $$x^3 - 6x^2 + 12x - 8 = (x-2)^3$$ 19. **Set equal to zero:** $$(x-2)^3 = 0 \implies x=2$$ 20. **Check domain restrictions:** Recall $$x \neq 0$$ and $$x \neq 2$$. Since $$x=2$$ is excluded, there is no solution. **Final answer:** The solution set is the empty set $$\varnothing$$.