Question

lf $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+3x-2=0$, then the equation whose roots are $\alpha (\beta +\gamma ),\beta (\gamma +\alpha ),\gamma (\alpha +\beta )$, is:

• A. $y^3+6y^2+9y+4=0$
• B. $y^3+6y^2-9y+4=0$
• C. $y^3+6y^2+9y-4=0$
• D. $y^3-6y^2+9y+4=0$

Answer 1

Answer: (D)

$y^3-6y^2+9y+4=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+3x-2=0$
We have
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\alpha \gamma =3s_3=\alpha \beta \gamma =2$
Let $y=\alpha \beta +\alpha \gamma \Rightarrow y+\beta \gamma =\alpha \beta +\beta \gamma +\alpha \gamma =3$
$\Rightarrow y+\frac{\alpha \beta \gamma }{\alpha }=3\Rightarrow y+\frac{2}{\alpha }=3\Rightarrow \frac{2}{\alpha }=3-y\Rightarrow \alpha =\frac{2}{3-y}$
Replacing $x\to \frac{2}{3-y}$ in given equation, we get
${\left(\frac{2}{3-y}\right)}^3+3\left(\frac{2}{3-y}\right)-2=0\Rightarrow 8+6{(3-y)}^2-2{(3-y)}^3=0\Rightarrow 8+6(9+y^2-6y)-2(27-y^3-27y+9y^2)=0\Rightarrow y^3-6y^2+9y+4=0$