1. Given equation and roots

The cubic is $x^3+bx+c=0$ with roots $\alpha,\beta,\gamma$.

For a monic cubic $x^3+0x^2+bx+c=0$, by Vieta: $\alpha+\beta+\gamma=0,$ $\alpha\beta+\beta\gamma+\gamma\alpha=b,$ $\alpha\beta\gamma=-c.$

Also given: $\beta\gamma=1=-\alpha \implies \alpha=-1,\quad \beta\gamma=1.$

2. Find $\beta+\gamma$

Using $\alpha+\beta+\gamma=0,$ we get $-1+\beta+\gamma=0 \implies \beta+\gamma=1.$

So $\beta,\gamma$ are numbers with $\beta+\gamma=1,\qquad \beta\gamma=1.$

3. Find $b$ and $c$

Using Vieta: $b=\alpha\beta+\beta\gamma+\gamma\alpha.$ Substitute $\alpha=-1$ and $\beta\gamma=1$: $b=-\beta+1-\gamma=1-(\beta+\gamma)=1-1=0.$

Next, $\alpha\beta\gamma=-c.$ Since $\alpha=-1$ and $\beta\gamma=1$, $(-1)(1)=-c \implies -1=-c \implies c=1.$

Thus, $b=0,\qquad c=1,\qquad \alpha=-1.$

4. Compute $\beta^3+\gamma^3$ and relate $\beta^3,\gamma^3$

From $\beta+\gamma=1,\quad \beta\gamma=1,$ $\beta^3+\gamma^3=(\beta+\gamma)^3-3\beta\gamma(\beta+\gamma)=1^3-3(1)(1)=1-3=-2.$

Also, $\beta$ and $\gamma$ are roots of $t^2-(\beta+\gamma)t+\beta\gamma=0\implies t^2-t+1=0.$ So each satisfies $t^2=t-1.$ Multiplying by $t$: $t^3=t^2-t=(t-1)-t=-1.$ Hence $\beta^3=\gamma^3=-1.$

5. Evaluate the expression

We need $b^3+2c^3-3\alpha^3-6\beta^3-8\gamma^3.$ Substitute $b=0,\quad c=1,\quad \alpha=-1,\quad \beta^3=-1,\quad \gamma^3=-1.$

Then $0^3+2(1)^3-3(-1)^3-6(-1)-8(-1).$ Now, $=0+2-3(-1)+6+8$ $=2+3+6+8=19.$

6. Option check

The value is $\boxed{19}$ which corresponds to Option B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.