Question

If $\alpha ,\beta ,\gamma$ are the roots of a cubic equation satisfying the relations $\alpha +\beta +\gamma =2$, ${\alpha }^2+{\beta }^2+{\gamma }^2=6$ and ${\alpha }^3+{\beta }^3+{\gamma }^3=8$, then the equation is:

• A. $x^3+2x^2-x+2=0$
• B. $x^3-2x^2-x+2=0$
• C. $x^3-2x^2+x+2=0$
• D. $x^3-3x^2-x+2=0$

Answer 1

Answer: (B)

$x^3-2x^2-x+2=0$

As $\alpha ,\beta ,\gamma$ are roots of $a{x}^3+b{x}^2+cx+d=0$
$S_1=\alpha +\beta +\gamma =2S_2=\frac{{(\alpha +\beta +\gamma )}^2-({\alpha }^2+{\beta }^2+{\gamma }^2)}{a}=\frac{4-6}{2}=-1S_3=\frac{({\alpha }^3+{\beta }^3+{\gamma }^3)-({\alpha }^2+{\beta }^2+{\gamma }^2)(\alpha +\beta +\gamma )+(\alpha +\beta +\gamma )(\alpha \beta +\beta \gamma +\gamma \alpha )}{3}$
$=\frac{8-\left(2\right)\left(6\right)+\left(2\right)(-1)}{3}=\frac{8-12-2}{3}=-2$
Therefore
$x^3-\left(S_1\right)x^2+\left(S_2\right)x+(-S_3)=0\Rightarrow x^3-2x^2-x+2=0$