Question

If $\alpha ,\beta ,\gamma$ are non zero roots of $x^3+p{x}^2+qx+r=0$, then the equation whose roots are $\alpha (\beta +\gamma ),\beta (\gamma +\alpha ),\gamma (\alpha +\beta )$

• A. $x^3-2q{x}^2+(pr+q^2)x+(r^2-pqr)=0$
• B. $x^3-2q{x}^2+(pr+q^2)x+(r^2+pqr)=0$
• C. $x^3+2q{x}^2+(pr+q^2)x+(r^2-pqr)=0$
• D. $x^3+2q{x}^2+(pr+q^2)x+(r^2+pqr)=0$

Answer 1

The correct option is A $x^3-2q{x}^2+(pr+q^2)x+(r^2-pqr)=0$

Given equation:
$x^3+p{x}^2+qx+r=0$
We know that,
$\alpha +\beta +\gamma =-p\alpha \beta +\beta \gamma +\alpha \gamma =q\alpha \beta \gamma =-rr\ne 0(\because \alpha ,\beta ,\gamma \ne 0)$
Let
$y=\alpha (\beta +\gamma )=q-\frac{\alpha \beta \gamma }{\alpha }=q-\frac{(-r)}{\alpha }$
$\Rightarrow y-q=\frac{r}{\alpha }\Rightarrow \alpha =\frac{r}{y-q}$
which is a root of given equation.

Therefore, the equation whose roots are $\alpha (\beta +\gamma ),\beta (\gamma +\alpha ),\gamma (\alpha +\beta )$ is,

$\frac{r^3}{(y-q{)}^3}+p{\left(\frac{r}{y-q}\right)}^2+q\left(\frac{r}{y-q}\right)+r=0$
$\Rightarrow r^3+p{r}^2(y-q)+qr(y-q{)}^2+r(y-q{)}^3=0$
$\Rightarrow y^3-2q{y}^2+(pr+q^2)y+(r^2-pqr)=0(\because r\ne 0)$

Hence, the transformed equation is
$x^3-2q{x}^2+(pr+q^2)x+(r^2-pqr)=0$