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

Question

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

Accepted Answer

$\beta\gamma=1$ and $\alpha=-1$. Substituting $\alpha=-1$ into $x^3+bx+c=0$: $-1-b+c=0 \implies c-b=1$.

Also $\alpha\beta\gamma=-c \implies (-1)(1)=-c \implies c=1$, so $b=0$.

The equation becomes $x^3+1=0$, with roots $\alpha=-1,\ \beta=-\omega,\ \gamma=-\omega^2$ (where $\omega$ is a primitive cube root of unity, so $\alpha^3=\beta^3=\gamma^3=-1$).

$$b^3+2c^3-3\alpha^3-6\beta^3-8\gamma^3 = 0+2-3(-1)-6(-1)-8(-1)=0+2+3+6+8=19.$$

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