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  4. If α, β, γ are the roots of the cubic x3-2 x+3=0 then the value of (1/α3+β3+6)+(1/β3+γ3+6)+(1/γ3+α3+6) equals

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Q. If $\alpha, \beta, \gamma$ are the roots of the cubic $x^3-2 x+3=0$ then the value of $\frac{1}{\alpha^3+\beta^3+6}+\frac{1}{\beta^3+\gamma^3+6}+\frac{1}{\gamma^3+\alpha^3+6}$ equals

Complex Numbers and Quadratic Equations

Solution:

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$\alpha+\beta+\gamma=0 ; \sum \alpha \beta=-2$ and $\alpha \beta \gamma=-3$
$\alpha^3-2 \alpha+3=0 \Rightarrow \alpha^3=2 \alpha-3 $ .....(1)
$\|^{1 y} \beta^3=2 \beta-3 \text { and }$ ....(2)
$\gamma^3=2 \gamma-3 $ ....(3)
$\therefore \alpha^3+\beta^3=2(\alpha+\beta)-6 $
$\alpha^3+\beta^3+6=2(\alpha+\beta)$
$\therefore \frac{1}{\alpha^3+\beta^3+6}=\frac{1}{2(\alpha+\beta)}$
$\therefore \sum \frac{1}{\alpha^3+\beta^3+6}=\frac{1}{2}\left[\frac{1}{\alpha+\beta+\gamma-\gamma}+\frac{1}{\beta+\gamma+\alpha-\alpha}+\frac{1}{\gamma+\alpha+\beta-\beta}\right]$
$=\frac{-1}{2}\left[\frac{1}{\gamma}+\frac{1}{\alpha}+\frac{1}{\beta}\right] (\alpha+\beta+\gamma=0) $
$=\frac{-1}{2}\left[\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}\right]=\frac{-1}{2}\left[\frac{-2}{-3}\right]=\frac{-1}{3}$