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(A) Given the cubic equation $x^3-13x^2+kx+189=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$1) \alpha+\beta+\gamma=13$$2) \alp

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$4: 3$

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(A) Given the cubic equation $x^3-13x^2+kx+189=0$ with roots $\alpha, \beta, \gamma$.From Vieta's formulas,we have:$1) \alpha+\beta+\gamma=13$$2) \alpha\beta+\beta\gamma+\gamma\alpha=k$$3) \alpha\beta\gamma=-189$Given $\beta-\gamma=2$,let $\beta+\gamma=S$ and $\beta\gamma=P$.Then $\beta = \frac{S+2}{2}$ and $\gamma = \frac{S-2}{2}$,so $P = \frac{S^2-4}{4}$.From $(1)$,$\alpha+S=13 \implies \alpha=13-S$.Substitute into $(3)$: $(13-S) 	imes \frac{S^2-4}{4} = -189 \implies (13-S)(S^2-4) = -756$.$13S^2-52-S^3+4S = -756 \implies S^3-13S^2-4S-704=0$.Testing values,if $S=11$,$1331-13(121)-4(11)-704 = 1331-1573-44-704 
eq 0$. If $S=12$,$1728-1872-48-704 
eq 0$. If $S=14$,$2744-2548-56-704 
eq 0$. Let's recheck: $\alpha=13-S$. $\alpha\beta\gamma = (13-S)(\frac{S^2-4}{4}) = -189 \implies (13-S)(S^2-4) = -756$.For $S=14$,$(13-14)(196-4) = -192 
eq -756$. For $S=16$,$(13-16)(256-4) = -3(252) = -756$. Thus $S=16$ is not correct. Wait,$S=16$ gives $-756$. So $\beta+\gamma=16$ and $\alpha=13-16=-3$.Then $k = \alpha(\beta+\gamma) + \beta\gamma = -3(16) + \frac{16^2-4}{4} = -48 + 60 = 12$.We need $\beta+\gamma : k+\alpha = 16 : (12-3) = 16 : 9$. Re-evaluating: $\alpha=-3$ is a root,so $(-3)^3-13(-3)^2+k(-3)+189=0 \implies -27-117-3k+189=0 \implies 45-3k=0 \implies k=15$.Then $\beta+\gamma = 13-(-3) = 16$ and $k+\alpha = 15-3 = 12$. Ratio $16:12 = 4:3$.

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3-13x^2+kx+189=0$ such that $\beta-\gamma=2$,then $\beta+\gamma: k+\alpha=$

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