1. **State the problem:** We have a curve $C$ defined by the implicit equation $$x^3 + 2xy - x - y^3 - 20 = 0$$ We are given the derivative expression $$\frac{dy}{dx} = \frac{3x^2 + 2y - 1}{3y^2 - 2x}$$ We need to find the equation of the tangent line to the curve at the point $(3, -2)$ in the form $ax + by + c = 0$ where $a,b,c$ are integers. 2. **Find the slope of the tangent line at $(3,-2)$:** Substitute $x=3$ and $y=-2$ into the derivative formula: $$\frac{dy}{dx} = \frac{3(3)^2 + 2(-2) - 1}{3(-2)^2 - 2(3)} = \frac{3 \times 9 - 4 - 1}{3 \times 4 - 6} = \frac{27 - 4 - 1}{12 - 6} = \frac{22}{6}$$ Simplify the fraction: $$\frac{22}{6} = \frac{\cancel{2} \times 11}{\cancel{2} \times 3} = \frac{11}{3}$$ So the slope $m = \frac{11}{3}$. 3. **Write the tangent line equation using point-slope form:** The point-slope form is $$y - y_1 = m(x - x_1)$$ Substitute $m=\frac{11}{3}$ and point $(3,-2)$: $$y - (-2) = \frac{11}{3}(x - 3)$$ $$y + 2 = \frac{11}{3}x - 11$$ 4. **Rewrite in standard form $ax + by + c = 0$ with integer coefficients:** Multiply both sides by 3 to clear the denominator: $$3(y + 2) = 11x - 33$$ $$3y + 6 = 11x - 33$$ Bring all terms to one side: $$11x - 3y - 33 - 6 = 0$$ $$11x - 3y - 39 = 0$$ 5. **Final answer:** The equation of the tangent line is $$11x - 3y - 39 = 0$$ This is in the form $ax + by + c = 0$ with $a=11$, $b=-3$, and $c=-39$.