Animated Solution for Mathematics - Quadratic Equations: 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

The given cubic equation is $x^3+bx+c=0$.

The absence of an $x^2$ term is a significant observation. According to Vieta's formulas, the sum of the roots $\alpha +\beta +\gamma$ is equal to the negative coefficient of the $x^2$ term. Since this term is missing, we conclude that:

We are provided with the constraint $\beta \gamma =1=-\alpha$.

From the equality $1=-\alpha$, we immediately determine the value of the first root:

Using the sum of roots relation $\alpha +\beta +\gamma =0$, we substitute $\alpha =-1$:

Next, we apply Vieta's formula for the product of the roots, $\alpha \beta \gamma =-c$. Substituting $\alpha =-1$ and $\beta \gamma =1$:

For the coefficient $b$, we use the sum of roots taken two at a time: $\alpha \beta +\beta \gamma +\gamma \alpha =b$. Factoring this expression, we get:

Substituting our known values $\alpha =-1$, $\beta +\gamma =1$, and $\beta \gamma =1$:

The cubic equation simplifies to $x^3+1=0$, which implies $x^3=-1$ for any root $x$. Consequently, ${\alpha }^3=-1$, ${\beta }^3=-1$, and ${\gamma }^3=-1$.

We now evaluate the expression $b^3+2c^3-3{\alpha }^3-6{\beta }^3-8{\gamma }^3$ using $b=0$, $c=1$, and the cubic values of the roots:

Simplifying the arithmetic:

The final result is 19.