Let α ,   β ,   γ be the three roots of the equation x 3 + b x + c = 0 if β…

Given that roots of $x^3+bx+c=0$ are $\alpha , \beta , \gamma$

Also given that

$\beta \gamma =1=-\alpha$

$\Rightarrow \alpha =-1...\left(\mathrm{i}\right)$

Now let us apply the relation between the roots of the cubic equation with the coefficients.

$\Rightarrow \alpha +\beta +\gamma =0...\left(\mathrm{ii}\right)$

$\Rightarrow \alpha \beta \gamma =-c...\left(\mathrm{iii}\right)$

$\Rightarrow \left(-1\right)\left(1\right)=-c$

$\therefore c=1...\left(\mathrm{iv}\right)$

On substituting the value of $\alpha =-1$ in eq $\left(ii\right)$ we get,

$\beta +\gamma =1...\left(\mathrm{v}\right)$

$\because \alpha \beta +\beta \gamma +\gamma \alpha =b$

$\Rightarrow \alpha \left(\beta +\gamma \right)+\beta \gamma =b$

$\therefore b=0...\left(\mathrm{vi}\right)$

Hence, equation will be,

$x^3+1=0$

Whose roots will be, $-1,-\omega , -{\omega }^2$

We know that $1+\omega +{\omega }^2=0$

$\Rightarrow \alpha +\beta +\gamma =0$

$\Rightarrow -1+\beta +\gamma =0$

$\therefore \beta =-\omega , \gamma =-{\omega }^2...\left(\mathrm{vii}\right)$

$\Rightarrow {\beta }^3=-{\omega }^3=-1$ and $\Rightarrow {\gamma }^3=-{\omega }^6=-1$

Hence the value of 

$b^3+2c^3-3{\alpha }^3-6{\beta }^3-8{\gamma }^3$ is

$=0+2+3+6+8$

$=19$

Hence this is the correct option.