Question

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

• A. $(y-q{)}^3=0$
• B. $(y-r{)}^3=0$
• C. $(y-q+r{)}^3=0$
• D. $(y+q-r{)}^3=0$

Answer 1

The correct option is A $(y-q{)}^3=0$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+qx+r=0$
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\alpha \gamma =q{s}_3=\alpha \beta \gamma =-r$

Now $\alpha \beta +\beta \gamma +\alpha \gamma =q\Rightarrow \alpha \alpha \beta +\alpha \beta \gamma +\alpha \alpha \gamma =q\alpha$
$\Rightarrow {\alpha }^2(\beta +\gamma )=q\alpha -\alpha \beta \gamma \Rightarrow -{\alpha }^3=q\alpha +r$ ...(1)

Let $y=\beta \gamma -{\alpha }^2\Rightarrow \alpha y=\alpha \beta \gamma -{\alpha }^3$

$\Rightarrow \alpha y=-r+q\alpha +r$

$\Rightarrow \alpha (y-q)=0$

As $\alpha \ne 0\Rightarrow (y-q)=0\Rightarrow {(y-q)}^3=0$