Let $\alpha,\beta,\gamma$ be roots of $x^3+bx+c=0$ with $\beta\gamma=1=-\alpha$. Then $b^3+2c^3-3\alpha^3-6\beta^3-8\gamma^3$ is equal to
Step-by-Step Solution
Key Concept: $\alpha=-1$, $\beta\gamma=1$, and $\alpha+\beta+\gamma=0\Rightarrow\beta+\gamma=1$. $b=\alpha\beta+\beta\gamma+\gamma\alpha=0$, $c=1$. Equation: $x^3+1=0$, roots $-1,-\omega,-\omega^2$.
$b=0,c=1,\alpha=-1,\beta^3=-1,\gamma^3=-1$. Expression $=19$.
Correct Answer: 4