Question

Given $\alpha +\beta +\gamma =2,{\alpha }^2+{\beta }^2+{\gamma }^2=6$ and ${\alpha }^3+{\beta }^3+{\gamma }^3=8$, then the equation whose roots are $\alpha ,\beta ,\gamma$ is:

• A. $x^3-2x^2+x-2=0$
• B. $x^3-2x^2-x+2=0$
• C. $x^3-2x^2+2x-4=0$
• D. $x^3-2x^2+3x-6=0$

Answer 1

Answer: (B)

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

We know $2(\alpha \beta +\beta \gamma +\gamma \alpha )={(\alpha +\beta +\gamma )}^2-({\alpha }^2+{\beta }^2+{\gamma }^2)=4-6=-2\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =-1$
And
${\alpha }^3+{\beta }^3+{\gamma }^3-3\alpha \beta \gamma =(\alpha +\beta +\gamma )({\alpha }^2+{\beta }^2+{\gamma }^2-\alpha \beta -\beta \gamma -\gamma \alpha )\Rightarrow 3\alpha \beta \gamma =8-2(6+1)=-6\Rightarrow \alpha \beta \gamma =-2$
So, the required equation is
$x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma =0\Rightarrow x^3-2x^2-x+2=0$