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Let \alpha, \beta, \gamma, be the three roots of the equation x^3+b x+c=0 If  \beta \gamma=1=-\alpha, then b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3 is

Option: 1

\frac{155}{8}


Option: 2

21


Option: 3

19


Option: 4

\frac{169}{8}


Answers (1)

best_answer

\begin{aligned} & \beta \gamma=1 \\ & \alpha=-1 \\ & \text { Put } \alpha=-1 \\ & -1-b+c=0 \\ & c-b=1\end{aligned}

also

\begin{aligned} & \alpha \cdot \beta \cdot \gamma=-\mathrm{c} \\ & -1=-\mathrm{c} \Rightarrow \mathrm{c}=1 \\ & \therefore \mathrm{b}=0 \\ & \mathrm{x}^3+1=0 \\ & \alpha=-1, \beta=-\mathrm{w}, \gamma=-\mathrm{w}^2 \\ & \therefore \mathrm{b}^3+2 \mathrm{c}^3-3 \alpha^3-6 \beta^3-8 \gamma^3 \\ & 0+2+3+6+8=19 \end{aligned}

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Pankaj Sanodiya

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