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 ${b}^{3}+2{c}^{3}-3{\alpha }^{3}-6{\beta }^{3}-8{\gamma }^{3}$ is equal to :

Accepted Answer

Given cubic equation is :

$x^3+b x+c=0$

$\because \alpha, \beta, \gamma$ are the roots of above equation.

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

$$
\begin{aligned}
& \text { So, product of roots }=-c \\
& \Rightarrow \alpha \beta \gamma=-c \\
& \Rightarrow(-1)(1)=-c \\
& \Rightarrow c=1
\end{aligned}
$$

Since, $\alpha=-1$ is the root. So,

$$
\begin{aligned}
& \Rightarrow-1-b+c=0 \\
& \Rightarrow c-b=1 \\
& \Rightarrow 1-b=1 \Rightarrow b=0
\end{aligned}
$$

The given equation becomes $x^3+1=0$

So, roots are $-1,-\omega,-\omega^2$

$$
\begin{aligned}
& \therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\
& =0+2-3(-1)^3-6(-\omega)^3-8\left(-\omega^2\right)^3 \\
& =2+3+6 \omega^3+8 \omega^6 \\
& =5+6+8=19
\end{aligned}
$$

Concept :

For a cubic equation, $a x^3+b x^2+c x+d=0$

Sum of roots $=\frac{-b}{a}$

Product of roots taken two at a time $=\frac{c}{a}$

Product of roots $=\frac{-d}{a}$

Solutions · 1