Let $$\alpha, \beta, \gamma$$ be the three roots of the equation $$x^3 + bx + c = 0$$ if $$\beta\gamma = 1 = -\alpha$$ then $$b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$$ is equal to

Consider the cubic equation $$x^3 + bx + c = 0$$ with roots $$\alpha,\beta,\gamma$$. The conditions $$\beta\gamma = 1$$ and $$-\alpha = 1$$ imply $$\alpha = -1$$.

Using Vieta's formulas, the relation $$\alpha + \beta + \gamma = 0$$ gives $$\beta + \gamma = 1$$. The sum of products $$\alpha\beta + \beta\gamma + \gamma\alpha = b$$ becomes $$-(\beta + \gamma) + 1 = b$$, so $$b = 0$$. Finally, the product $$\alpha\beta\gamma = -c$$ gives $$(-1)(1) = -c$$, leading to $$c = 1$$.

Since $$\beta + \gamma = 1$$ and $$\beta\gamma = 1$$, the numbers $$\beta$$ and $$\gamma$$ satisfy the quadratic equation $$t^2 - t + 1 = 0$$. It follows that $$\beta = \frac{1 + i\sqrt{3}}{2} = e^{i\pi/3}$$ and $$\gamma = \frac{1 - i\sqrt{3}}{2} = e^{-i\pi/3}$$.

Raising these roots to the third power yields $$\beta^3 = e^{i\pi} = -1$$ and $$\gamma^3 = e^{-i\pi} = -1$$.

Substituting into the expression $$b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3$$ yields
$$0 + 2(1) - 3(-1)^3 - 6(-1) - 8(-1) = 0 + 2 + 3 + 6 + 8 = 19\,. $$