Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3+p{x}^2+qx+r=0,r\ne 0$ and $\beta \gamma +\frac{1}{\alpha },\alpha \gamma +\frac{1}{\beta },\alpha \beta +\frac{1}{\gamma }$ are the roots of the equation $x^3+a{x}^2+bx+c=0,c\ne 0,$ then the value of $c$ is

• A. $(1-r{)}^3$
• B. $(r-1{)}^3$
• C. $\frac{(1-r{)}^3}{r}$
• D. $\frac{(r-1{)}^3}{r}$

Answer 1

The correct option is C $\frac{(1-r{)}^3}{r}$

$x^3+p{x}^2+qx+r=0...\left(1\right)$
$x^3+a{x}^2+bx+c=0...\left(2\right)$
Let $m$ is a root of eqn$\left(1\right)$ and $n$ is a root of eqn$\left(2\right)$.
$n=\beta \gamma +\frac{1}{\alpha }=\frac{\alpha \beta \gamma +1}{\alpha }=\frac{1-r}{\alpha }[\because \alpha \beta \gamma =-r]$
$\Rightarrow n=\frac{1-r}{m}\Rightarrow m=\frac{1-r}{n}$
Substitute $m=\frac{1-r}{n}$ in eqn$\left(1\right)$, we get
${\left(\frac{1-r}{n}\right)}^3+p{\left(\frac{1-r}{n}\right)}^2+q\left(\frac{1-r}{n}\right)+r=0$
$\Rightarrow r{n}^3+q(1-r)n^2+p(1-r{)}^{2}n+(1-r{)}^3=0$
Replacing $n$ by $x$, we get
$x^3+\frac{q(1-r)}{r}{x}^2+\frac{p(1-r{)}^2}{r}x+\frac{(1-r{)}^3}{r}=0$
$\therefore c=\frac{(1-r{)}^3}{r}$