Question

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

• A. $y^3-9y^2-24y-16=0$
• B. $y^3-9y^2+24y-16=0$
• C. $y^3+9y^2+24y-16=0$
• D. $y^3+9y^2+24y+16=0$

Answer 1

The correct option is A $y^3-9y^2-24y-16=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+3x-2=0$
we have $s_3=\alpha \beta \gamma =2$
Let $y={\beta }^2{\gamma }^2\Rightarrow {\alpha }^{2}y={\alpha }^2{\beta }^2{\gamma }^2=4\Rightarrow {\alpha }^2=\frac{4}{y}\alpha =\sqrt{\frac{4}{y}}$
Replace $x\to \sqrt{\frac{4}{y}}$ in given equation
${\left(\sqrt{\frac{4}{y}}\right)}^3+3\left(\sqrt{\frac{4}{y}}\right)-2=0$

$\Rightarrow 8+6y-2y\sqrt{y}=0\Rightarrow 2y\sqrt{y}=8+6y$

$\Rightarrow {\left(y\sqrt{y}\right)}^2={(4+3y)}^2\Rightarrow y^3=16+9y^2+24y$

$\Rightarrow y^3-9y^2-24y-16=0$