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

Given cubic equation is : <br/><br/>$x^3+b x+c=0$ <br/><br/>$\because \alpha, \beta, \gamma$ are the roots of above equation. <br/><br/>And $\beta \gamma=1=-\alpha$ <br/><br/>$$ \begin{aligned} & \text { So, product of roots }=-c \\\\ & \Rightarrow \alpha \beta \gamma=-c \\\\ & \Rightarrow(-1)(1)=-c \\\\ & \Rightarrow c=1 \end{aligned} $$ <br/><br/>Since, $\alpha=-1$ is the root. So, <br/><br/>$$ \begin{aligned} & \Rightarrow-1-b+c=0 \\\\ & \Rightarrow c-b=1 \\\\ & \Rightarrow 1-b=1 \Rightarrow b=0 \end{aligned} $$ <br/><br/>The given equation becomes $x^3+1=0$ <br/><br/>So, roots are $-1,-\omega,-\omega^2$ <br/><br/>$$ \begin{aligned} & \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\\\ & =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\\\ & =2+3+6 \omega^3+8 \omega^6 \\\\ & =5+6+8=19 \end{aligned} $$ <br/><br/><b>Concept :</b> <br/><br/>For a cubic equation, $a x^3+b x^2+c x+d=0$ <br/><br/>Sum of roots $=\frac{-b}{a}$ <br/><br/>Product of roots taken two at a time $=\frac{c}{a}$ <br/><br/>Product of roots $=\frac{-d}{a}$