Question

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

• A. $y^3+6q{y}^2+9q^{2}y+({4q}^3+{27r}^2)=0$
• B. $y^3+6q{y}^2+9q^{2}y+({4r}^3-{27r}^2)=0$
• C. $y^3+6q{y}^2+9q^{2}y+({4r}^3+{27q}^2)=0$
• D. $y^3+6q{y}^2+9q^{2}y+({4q}^3-{27r}^2)=0$

Answer 1

The correct option is A $y^3+6q{y}^2+9q^{2}y+({4q}^3+{27r}^2)=0$

$x^3+qx+r=0$ has $\alpha ,\beta ,\gamma$ as roots.
$\therefore$ Sum of roots: $\alpha +\beta +\gamma =0$
Product of roots taken two at a time: $\alpha \beta +\beta \gamma +\gamma \alpha =q$
Product of roots: $\alpha \beta \gamma =-r$
${\alpha }^3+q\alpha +r=0$...............(i)
${\beta }^3+q\beta +r=0$............(ii)
Now, (i) - (ii), we get
${\alpha }^3-{\beta }^3+q(\alpha -\beta )=0$
$\Rightarrow (\alpha -\beta )({\alpha }^2+{\beta }^2+\alpha \beta +q)=0$
$\Rightarrow (\alpha -\beta )\left(\right(\alpha -\beta {)}^2+3\alpha \beta +q)=0$....................... {1}
Similarly
$\Rightarrow (\beta -\gamma )\left(\right(\beta -\gamma {)}^2+3\beta \gamma +q)=0$....................... {2}
$\Rightarrow (\gamma -\alpha )\left(\right(\gamma -\alpha {)}^2+3\gamma \alpha +q)=0$....................... {3}
Clearly, $\alpha \ne \beta \ne \gamma$
$\therefore$ From {1}, {2} and {3}, we get
$(\alpha -\beta {)}^2=-3\alpha \beta -q$
$(\beta -\gamma {)}^2=-3\beta \gamma -q$
$(\gamma -\alpha {)}^2=-3\gamma \alpha -q$
Now, we have to find the equation with roots $(\alpha -\beta {)}^2,(\beta -\gamma {)}^2,(\gamma -\alpha {)}^2$
Sum of roots:
$(\alpha -\beta {)}^2+(\beta -\gamma {)}^2+(\gamma -\alpha {)}^2=-3(\alpha \beta +\beta \gamma +\gamma \alpha )-3q=-6q$
Product of roots taken two at a time:
$(\alpha -\beta {)}^2(\beta -\gamma {)}^2+(\beta -\gamma {)}^2(\gamma -\alpha {)}^2+(\gamma -\alpha {)}^2(\alpha -\beta {)}^2$
$=(3\alpha \beta +q)(3\beta \gamma +q)+(3\beta \gamma +q)(3\gamma \alpha +q)+(3\gamma \alpha +q)(3\alpha \beta +q)=9q^2$
Product of roots:
$\left(\right(\alpha -\beta \left)\right(\beta -\gamma \left)\right(\gamma -\alpha ){)}^2=(3\alpha \beta +q\left)\right(3\beta \gamma +q\left)\right(3\gamma \alpha +q)$
$=9q\alpha \beta \gamma (\alpha +\beta +\gamma )+q^3+3q^2(\alpha \beta +\beta \gamma +\gamma \alpha )+27(\alpha \beta \gamma {)}^2=4q^3+27r^2$
$\therefore$ the equation is $y^3+6q{y}^2+9q^{2}y+({4q}^3+{27r}^2)=0$.