(B) Given equation $x^3-6x^2+11x-6=0$ has the roots $\alpha, \beta, \gamma$. By factoring,$(x-1)(x-2)(x-3)=0$,so the roots are $1, 2, 3$. Let $\alpha=

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(B) Given equation $x^3-6x^2+11x-6=0$ has the roots $\alpha, \beta, \gamma$. By factoring,$(x-1)(x-2)(x-3)=0$,so the roots are $1, 2, 3$. Let $\alpha=1, \beta=2, \gamma=3$. Then,$a = \alpha^2+\beta^2+\gamma^2 = 1^2+2^2+3^2 = 1+4+9 = 14$. $b = \alpha\beta+\beta\gamma+\gamma\alpha = (1)(2)+(2)(3)+(3)(1) = 2+6+3 = 11$. $c = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (1+2)(2+3)(3+1) = 3 \times 5 \times 4 = 60$. Comparing the values,$11 < 14 < 60$,which implies $b < a < c$.

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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x-6=0$ and if $a=\alpha^2+\beta^2+\gamma^2$,$b=\alpha\beta+\beta\gamma+\gamma\alpha$ and $c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$,then the correct inequality among the following is

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$a < b < c$
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$b < a < c$
C
$b < c < a$
D
$c < a < b$

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If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x-6=0$ and if $a=\alpha^2+\beta^2+\gamma^2$,$b=\alpha\beta+\beta\gamma+\gamma\alpha$ and $c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$,then the correct inequality among the following is

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