Given equation $x^3-6 x^2+11 x-6=0$
has the roots $\alpha, \beta \gamma$.
Given, $\quad a=\alpha^2+\beta^2+\gamma^2$ ...(i)
$b=\alpha \beta+\beta \gamma+\gamma \alpha$ ...(ii)
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$ ...(iii)
In cubic equation the sum of the roots
$\alpha+\beta+\gamma=-\left(\frac{-6}{1}\right)=6$
$\alpha \beta+\beta \gamma+\gamma \alpha=\left(\frac{11}{1}\right)=11$
product of the roots
$\alpha \cdot \beta \cdot \gamma=-\left(\frac{-6}{1}\right)=6$
From Eq. (ii), $b=11$
From Eq. (i), $a=\alpha^2+\beta^2+\gamma^2$
$\Rightarrow \quad a=(\alpha+\beta+\gamma)^2-2(\alpha \beta+\beta \gamma+\gamma \alpha)$
$\Rightarrow \quad a=(6)^2-2(11) \Rightarrow 36-22$
$\Rightarrow \quad a=14$
From Eq. (iii)
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$
$=\left(\alpha \beta+\beta^2+\alpha \gamma+\beta \gamma\right)(\gamma+\alpha)$
$=\alpha \beta \gamma+\beta^2 \gamma+\alpha \gamma^2+\beta \gamma^2+\alpha^2 \beta$ $+\alpha \beta^2+\alpha^2 \gamma+\alpha \beta \gamma$
$\Rightarrow c=[(\alpha+\beta+\gamma)-\gamma][(\alpha+\beta+\gamma)-\alpha]$ $[(\alpha+\beta+\gamma)-\beta]$
$=(6-\gamma)(6-\alpha)(6-\beta)$
$=(36-6 \gamma-6 \alpha+\alpha \gamma)(6-\beta)$
$=216-36 \gamma-36 \alpha+6 \alpha \gamma-36 \beta+6 \gamma \beta$ $+6 \alpha \beta-\alpha \beta \gamma$
$=216-\alpha \beta \gamma+6(\alpha \beta+\beta \gamma+\gamma \alpha)-36(\alpha+\beta+\gamma)$
$=216-6+6(11)-36(6)$
$=210+66-216=60$
Hence, $\quad c=60$
$\Rightarrow \quad b < a < c$Alternate method
Given equation is $x^3-6 x^2+11 x-6=0$
$\Rightarrow \quad(x-1)(x-2)(x-3)=0$
The roots of this equation are 1,2 , and 3 .
Let $\alpha=1, \beta=2, \gamma=3$
Now, $\quad a=\alpha^2+\beta^2+\gamma^2$
$=1+4+9=14$
$b=\alpha \beta+\beta \gamma+\gamma \alpha$
$=2+6+3=11$
$c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$
$=3 \cdot 5 \cdot 4=60$
From the values of $a, b$ and $c$ it is clear that $b < a < c$.