If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+a^2+b x$ $+c=0$, then the roots of the…
If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+a^2+b x$ $+c=0$, then the roots of the equation $x^3+\left(2 b-a^2\right) x^2+$ $\left(b^2-2 a c\right) x-c^2=0$ are
$\alpha^3, \beta^3, \gamma^3$
$(\alpha+1)^2,(\beta+1)^2,(\gamma+1)^2$
$\alpha^2, \beta^2, \gamma^2$
$(\alpha-1)^2,(\beta-1)^2,(\gamma-1)^2$
Solution
$\because \alpha, \beta$ and $\gamma$ are roots of $x^3+a^2+b x+c=0$
$\therefore \alpha+\beta+\gamma=-a \Rightarrow a=-(\alpha+\beta+\gamma)$
$\begin{aligned} & \alpha \beta+\beta \gamma+\gamma \alpha=b \\ & \alpha \beta \gamma=-c \Rightarrow c=-\alpha \beta \gamma\end{aligned}$
Let $\alpha^{\prime}, \beta^{\prime}$ and $\alpha^{\prime}$ are the roots of the equation
$\Rightarrow \alpha^{\prime}+\beta^{\prime}+\gamma^{\prime}=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$\Rightarrow \alpha^{\prime}+\beta^{\prime}+\gamma^{\prime}=\alpha^2+\beta^2+\gamma^2$ ...(i)
And $\alpha^{\prime} \beta^{\prime}+\beta^{\prime} \gamma^{\prime}+\gamma^{\prime} \alpha^{\prime}=b^2-2 a c$
$\begin{aligned} & \Rightarrow \alpha^{\prime} \beta^{\prime}+\beta^{\prime} \gamma^{\prime}+\gamma^{\prime} \alpha^{\prime}=(\alpha \beta+\beta \gamma+\gamma \alpha)^2-2(\alpha \beta \gamma) \alpha \beta \gamma \\ & \Rightarrow \alpha^{\prime} \beta^{\prime}+\beta^{\prime} \gamma^{\prime}+\gamma^{\prime} \alpha^{\prime}=(\alpha \beta)^2+(\beta \gamma)^2+\gamma^2 \alpha^2\end{aligned}$
$=\alpha^2 \beta^2+\beta^2 \gamma^2+\gamma^2 \alpha^2$
...(ii)
And $\alpha^{\prime} \beta^{\prime} \gamma^{\prime}=-\left(-c^2\right)=c^2=\alpha^2 \beta^2 \gamma^2$ ... (iii)
from eqs. (i), (ii) \& (iii) $\alpha^{\prime}=\alpha^2, \beta^{\prime}=\beta^2, \gamma^{\prime}=\gamma^2$
$\therefore$ Roots are $\alpha^2, \beta^2$ and $\gamma^2$