Question

lf $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3-x+2=0$, then the equation whose roots are $\alpha \beta +\frac{1}{\gamma },\beta \gamma +\frac{1}{\alpha },\gamma \alpha +\frac{1}{\beta }$, is:

• A. $2y^3+y^2+1=0$
• B. $2y^3-y^2+1=0$
• C. $y^3+y^2+1=0$
• D. $2y^3+y^2-1=0$

Answer 1

Answer: (D)

$2y^3+y^2-1=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3-x+2=0$
We get $s_3=\alpha \beta \gamma =-2$ ..(1)
Now $y=\alpha \beta +\frac{1}{\gamma }\Rightarrow y=\frac{\alpha \beta \gamma +1}{\gamma }$
Using (1)
$y=\frac{-2+1}{\gamma }\Rightarrow y=-\frac{1}{x}\Rightarrow x=-\frac{1}{y}$
Replacing $x\to -\frac{1}{y}$ in $x^3-x+2=0$ , we get
${(-\frac{1}{y})}^3-(-\frac{1}{y})+2=0\Rightarrow -1+y^2+2y^3=0\Rightarrow 2y^3+y^2-1=0$