Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+ax+b=0$, then the transformed equation having the roots $(\beta -\gamma {)}^2,(\gamma -\alpha {)}^2,(\alpha -\beta {)}^2$ is obtained by taking $x=$

• A. $\frac{b}{y+a}$
• B. $\frac{2b}{y+a}$
• C. $\frac{3b}{y+a}$
• D. $\frac{4b}{y+a}$

Answer 1

Answer: (C)

$\frac{3b}{y+a}$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+ax+b=0$
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\alpha \gamma =a{s}_3=\alpha \beta \gamma =-b$
$\alpha \beta +\beta \gamma +\alpha \gamma =a\Rightarrow \alpha \beta \gamma +\beta {\gamma }^2+\alpha {\gamma }^2=a\gamma \Rightarrow {\gamma }^2(\alpha +\beta )=a\gamma -\alpha \beta \gamma \Rightarrow -{\gamma }^3=a\gamma +b\Rightarrow {\gamma }^3=-a\gamma -b$
Let $y={(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta ={\gamma }^2-4\alpha \beta$
$\Rightarrow y\gamma ={\gamma }^3-4\alpha \beta \gamma =-a\gamma -b+4b=3b-a\gamma$
$\Rightarrow \gamma (y+a)=3b\Rightarrow \gamma =\frac{3b}{y+a}\Rightarrow x=\frac{3b}{y+a}$