Question

If $\alpha ,\beta and\gamma$ are the roots of the equation $x^3+3x+2=\mathrm{0...}\left(i\right)\text{and}(\alpha -\beta )(\alpha -\gamma ),(\beta -\gamma )(\beta -\alpha ),(\gamma -\alpha )(\gamma -\beta )$ are the roots of equation $y^3-9y^2-216=0...\left(ii\right)$.

Relation to transform the eq.(i) to eq.(ii) is given by .

• A. $y={\alpha }^2-\frac{1}{\alpha }$
• B. $y=2{\alpha }^2-\frac{1}{\alpha }$
• C. $y=2{\alpha }^2-\frac{2}{\alpha }$
• D. $y={\alpha }^2-\frac{2}{\alpha }$

Answer 1

The correct option is C

$y=2{\alpha }^2-\frac{2}{\alpha }$

$\alpha ,\beta ,\gamma$ are the roots of equation $x^3+3x+2=0$ ___________(1)

Using sum of the roots
$\alpha +\beta +\gamma =0\Rightarrow \beta +\gamma =-\alpha$

Sum of the roots taking two at a time $\alpha \beta +\beta \gamma +\gamma \alpha =3$

Product of the roots
$\alpha \beta \gamma =-2\Rightarrow \beta \gamma =\frac{-2}{\alpha }$

We can take $y=(\alpha -\beta )(\alpha -\gamma )$$={\alpha }^2-\alpha \beta -\alpha \gamma +\beta \gamma$

$={\alpha }^2-\alpha (\beta +\gamma )+\beta \gamma \beta +\gamma =-\alpha ,\beta \gamma =\frac{-2}{\alpha }$

= ${\alpha }^2-\alpha (-\alpha )+\frac{-2}{\alpha }$

$\Rightarrow y=2{\alpha }^2-\frac{2}{\alpha }$