1. **Problem statement:** Given a cubic equation $$x^3 + ax^2 + bx + c = 0$$ with roots $$\alpha, \beta, \gamma \in \mathbb{C}$$ such that $$\alpha + \beta, \beta + \gamma, \gamma + \alpha \neq 0,$$ find the cubic equation whose roots are $$\frac{\gamma}{\alpha + \beta}, \frac{\alpha}{\beta + \gamma}, \frac{\beta}{\gamma + \alpha}.$$\n\n2. **Recall Viète's formulas:** For roots $$\alpha, \beta, \gamma$$ of the cubic, we have:\n$$\alpha + \beta + \gamma = -a,$$\n$$\alpha\beta + \beta\gamma + \gamma\alpha = b,$$\n$$\alpha\beta\gamma = -c.$$\n\n3. **Express the new roots:** Define new roots as $$x_1 = \frac{\gamma}{\alpha + \beta}, x_2 = \frac{\alpha}{\beta + \gamma}, x_3 = \frac{\beta}{\gamma + \alpha}.$$\nNote that $$\alpha + \beta = -\gamma - a,$$ but more straightforward is to use the identity $$\alpha + \beta = -a - \gamma,$$ similarly for others.\n\n4. **Rewrite each root:** For example, $$x_1 = \frac{\gamma}{\alpha + \beta} = \frac{\gamma}{-a - \gamma}.$$ Similarly, $$x_2 = \frac{\alpha}{-a - \alpha}, x_3 = \frac{\beta}{-a - \beta}.$$\n\n5. **Set $$y = x_i$$ and solve for $$\alpha, \beta, \gamma$$ in terms of $$y$$:** From $$y = \frac{r}{-a - r}$$ where $$r$$ is one of the roots $$\alpha, \beta, \gamma,$$ we get:\n$$y(-a - r) = r \implies -ay - ry = r \implies r + ry = -ay \implies r(1 + y) = -ay \implies r = \frac{-ay}{1 + y}.$$\n\n6. **Since $$r$$ is a root of the original cubic, substitute $$r = \frac{-ay}{1 + y}$$ into the original polynomial:**\n$$\left(\frac{-ay}{1 + y}\right)^3 + a \left(\frac{-ay}{1 + y}\right)^2 + b \left(\frac{-ay}{1 + y}\right) + c = 0.$$\nMultiply both sides by $$(1 + y)^3$$ to clear denominators:\n$$(-ay)^3 + a(-ay)^2(1 + y) + b(-ay)(1 + y)^2 + c(1 + y)^3 = 0.$$\n\n7. **Expand and simplify:**\n$$-a^3 y^3 + a a^2 y^2 (1 + y) - b a y (1 + y)^2 + c (1 + y)^3 = 0.$$\nRewrite carefully:\n$$-a^3 y^3 + a^3 y^2 (1 + y) - b a y (1 + y)^2 + c (1 + y)^3 = 0.$$\n\n8. **Expand terms:**\n- $$a^3 y^2 (1 + y) = a^3 y^2 + a^3 y^3,$$\n- $$(1 + y)^2 = 1 + 2y + y^2,$$\n- $$(1 + y)^3 = 1 + 3y + 3y^2 + y^3.$$\n\n9. **Substitute expansions:**\n$$-a^3 y^3 + a^3 y^2 + a^3 y^3 - b a y (1 + 2y + y^2) + c (1 + 3y + 3y^2 + y^3) = 0.$$\n\n10. **Combine like terms:**\n$$(-a^3 y^3 + a^3 y^3) + a^3 y^2 - b a y - 2 b a y^2 - b a y^3 + c + 3 c y + 3 c y^2 + c y^3 = 0,$$\nwhich simplifies to\n$$a^3 y^2 - b a y - 2 b a y^2 - b a y^3 + c + 3 c y + 3 c y^2 + c y^3 = 0.$$\n\n11. **Group terms by powers of $$y$$:**\n- $$y^3: (-b a + c) y^3,$$\n- $$y^2: (a^3 - 2 b a + 3 c) y^2,$$\n- $$y: (-b a + 3 c) y,$$\n- constant: $$c.$$\n\n12. **Final cubic equation in $$y$$:**\n$$(-b a + c) y^3 + (a^3 - 2 b a + 3 c) y^2 + (-b a + 3 c) y + c = 0.$$\n\n13. **Summary:** The cubic equation with roots $$\frac{\gamma}{\alpha + \beta}, \frac{\alpha}{\beta + \gamma}, \frac{\beta}{\gamma + \alpha}$$ is\n$$(-b a + c) y^3 + (a^3 - 2 b a + 3 c) y^2 + (-b a + 3 c) y + c = 0.$$