Let α, β, γ be the three roots of the equation x ^ 3 + bx + c = 0...

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

Also given that

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

$⇒ \alpha = - 1 . . . ( i )$

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

$⇒ \alpha + \beta + \gamma = 0 . . . ( ii )$

$⇒ \alpha\beta\gamma = - c . . . ( iii )$

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

$∴ c = 1 . . . ( iv )$

On substituting the value of $\alpha = - 1$ in eq \left(ii\right)$we get,</p><p>$\beta + \gamma = 1 . . . ( v )$</p><p>$∵ \alpha\beta + \beta\gamma + \gamma\alpha = b$</p><p>$⇒ \alpha\left(\beta + \gamma\right) + \beta\gamma = b$</p><p>$∴ b = 0 . . . ( vi )$</p><p>Hence, equation will be,</p><p>${x}^{3} + 1 = 0$</p><p>Whose roots will be,$- 1, - \omega, - {\omega}^{2}$</p><p>We know that$1 + \omega + {\omega}^{2} = 0$</p><p>$⇒ \alpha + \beta + \gamma = 0$</p><p>$⇒ - 1 + \beta + \gamma = 0$</p><p>$∴ \beta = - \omega, \gamma = - {\omega}^{2} . . . ( vii )$</p><p>$⇒ {\beta}^{3} = - {\omega}^{3} = - 1$and$⇒ {\gamma}^{3} = - {\omega}^{6} = - 1$</p><p>Hence the value of </p><p>${b}^{3} + 2{c}^{3} - 3{\alpha}^{3} - 6{\beta}^{3} - 8{\gamma}^{3} is

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

$= 19$

Hence this is the correct option.