Let P(x) = x^3 - ax^2 + bx + c where a, b, c | vedclass

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(B) Given $P(x) = x^3 - ax^2 + bx + c$ has integral roots $\alpha, \beta, \gamma$.Thus,$P(x) = (x - \alpha)(x - \beta)(x - \gamma)$.Given $P(6) = 3$,w

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$15$

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(B) Given $P(x) = x^3 - ax^2 + bx + c$ has integral roots $\alpha, \beta, \gamma$.Thus,$P(x) = (x - \alpha)(x - \beta)(x - \gamma)$.Given $P(6) = 3$,we have $(6 - \alpha)(6 - \beta)(6 - \gamma) = 3$.Since $\alpha, \beta, \gamma$ are integers,$(6 - \alpha), (6 - \beta),$ and $(6 - \gamma)$ must be integer factors of $3$.The factors of $3$ are $\{1, -1, 3, -3\}$.Let $x_1 = 6 - \alpha, x_2 = 6 - \beta, x_3 = 6 - \gamma$. Then $x_1 x_2 x_3 = 3$.Also,$a = \alpha + \beta + \gamma = (6 - x_1) + (6 - x_2) + (6 - x_3) = 18 - (x_1 + x_2 + x_3)$.Possible sets for $(x_1, x_2, x_3)$ such that their product is $3$:$1)$ $(1, 1, 3) \Rightarrow x_1 + x_2 + x_3 = 5 \Rightarrow a = 18 - 5 = 13$.$2)$ $(1, -1, -3) \Rightarrow x_1 + x_2 + x_3 = -3 \Rightarrow a = 18 - (-3) = 21$.$3)$ $(-1, -1, 3) \Rightarrow x_1 + x_2 + x_3 = 1 \Rightarrow a = 18 - 1 = 17$.Comparing with the options,$a$ cannot be $15$.

Let $P(x) = x^3 - ax^2 + bx + c$ where $a, b, c \in R$ has integral roots such that $P(6) = 3$,then '$a$' cannot be equal to

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