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(C) Given the quadratic equation $x^2+bx+c=0$,the sum of roots is $\alpha+\beta = -b = 5$,which implies $b = -5$. Using the identity $\alpha^3+\beta^3

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$-3$

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(C) Given the quadratic equation $x^2+bx+c=0$,the sum of roots is $\alpha+\beta = -b = 5$,which implies $b = -5$. Using the identity $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta)$,we substitute the given values: $60 = (5)^3 - 3\alpha\beta(5)$. $60 = 125 - 15\alpha\beta$. $15\alpha\beta = 125 - 60 = 65$. $\alpha\beta = \frac{65}{15} = \frac{13}{3}$. Since $\alpha\beta = c$,we have $c = \frac{13}{3}$. Now,calculate $3c+2$: $3(\frac{13}{3}) + 2 = 13 + 2 = 15$. Since $b = -5$,we check the options: $2b = 2(-5) = -10$. $3b = 3(-5) = -15$. $-3b = -3(-5) = 15$. $-2b = -2(-5) = 10$. Thus,$3c+2 = -3b$.

If $\alpha, \beta$ are the roots of the equation $x^2+bx+c=0$ satisfying the conditions $\alpha+\beta=5$ and $\alpha^3+\beta^3=60$,then $3c+2=$ (in $b$)

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