If $\alpha, \beta, \gamma$ are the roots of $x^3-6 x^2+11 x-6=0$, then the equation having the roots…
If $\alpha, \beta, \gamma$ are the roots of $x^3-6 x^2+11 x-6=0$, then the equation having the roots $\alpha^2+\beta^2, \beta^2+\gamma^2$ and $\gamma^2+\alpha^2$ is
$x^3-28 x^2+245 x-650=0$
$x^3-28 x^2+245 x+650=0$
$x^3+28 x^2-245 x-650=0$
$x^3+28 x^2+245 x-650=0$
Solution
$
\begin{aligned}
& \text { } x^3-6 x^2+11 x-6=0 \\
& \Rightarrow(x-1)(x-2)(x-3)=0 \\
& \Rightarrow x=1,2,3
\end{aligned}
$
$\because \alpha, \beta, \gamma$ are the roots of the Eq.(i), so
$
\alpha=1, \beta=2, \gamma=3
$
Therefore, $\alpha^2+\beta^2=(1)^2+(2)^2=5=\alpha^{\prime}$ (say)
$
\beta^2+\gamma^2=(2)^2+(3)^2=13=\beta^{\prime}(\text { say })
$
and $\quad \gamma^2+\alpha^2=(3)^2+1=10=\gamma^{\prime}$ (say)
Equation of the having the roots $\alpha^{\prime}, \beta^{\prime}$ and $\gamma^{\prime}$,
$
\begin{aligned}
& x^3-\left(\alpha^{\prime}+\beta^{\prime}+\gamma^{\prime}\right) x^2+\left(\alpha^{\prime} \beta^{\prime}+\beta^{\prime} \gamma^{\prime}+\gamma^{\prime} \alpha^{\prime}\right) x \\
& -\alpha^{\prime} \beta^{\prime} \gamma^{\prime}=0 \\
& \Rightarrow x^3-(5+13+10) x^2+(5 \times 13+13 \times 10+10 \times 5) x \\
& -5 \times 13 \times 10=0 \\
& \Rightarrow x^3-28 x^2+245 x-650=0 \\
&
\end{aligned}
$