Question

If $\alpha ,\beta ,\gamma$ are the roots of $x^3+3x^2+2=0$ then find the equation whose roots are $\frac{\alpha }{\beta +\gamma },\frac{\beta }{\gamma +\alpha },\frac{\gamma }{\alpha +\beta }$

• A. $2x^3+33x^2-6x-2=0$
• B. $2x^3+33x^2+6x-2=0$
• C. $2x^3+33x^2-6x+2=0$
• D. $2x^3+33x^2+6x+2=0$

Answer 1

The correct option is D $2x^3+33x^2+6x+2=0$

Given $\alpha +\beta +\gamma =-3,\alpha \beta +\beta \gamma +\gamma \alpha =0,\alpha \beta \gamma =-2$
Let
$y=\frac{\alpha }{\beta +\gamma }=\frac{\alpha }{(\alpha +\beta +\gamma )-\alpha }=\frac{\alpha }{-3-\alpha }$
$\Rightarrow \alpha =\frac{-3y}{y+1}$
which is a root of the given equation.
$\therefore {\left(\frac{-3y}{y+1}\right)}^3+3{\left(\frac{-3y}{y+1}\right)}^2+2=0$
$\Rightarrow -27y^3+27y^2(y+1)+2(y+1{)}^3=0$
$\Rightarrow -27y^3+27y^3+27y^2+2y^3+6y^2+6y=0$
$\Rightarrow 2y^3+33y^2+6y+2=0$
$\therefore$ The required equation is $2x^3+33x^2+6x+2=0$