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(B) Given equation $x^3-6x^2+11x-6=0$ has roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = 6$ $\alpha\beta+\beta\gamma+\gamm

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$b < a < c$

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(B) Given equation $x^3-6x^2+11x-6=0$ has roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = 6$ $\alpha\beta+\beta\gamma+\gamma\alpha = 11$ $\alpha\beta\gamma = 6$ Calculating $a, b, c$: $b = \alpha\beta+\beta\gamma+\gamma\alpha = 11$ $a = \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 6^2 - 2(11) = 36 - 22 = 14$ $c = (\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (6-\gamma)(6-\alpha)(6-\beta)$ Since $x^3-6x^2+11x-6 = (x-1)(x-2)(x-3)$,the roots are $1, 2, 3$. $c = (6-1)(6-2)(6-3) = 5 	imes 4 	imes 3 = 60$ Comparing the values: $b=11, a=14, c=60$. Thus,$b < a < c$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

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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