Let α, β, gamma be three roots of equation x ^3+ bx + c =0 . If β gamma=1=-α , then b^3+2 c^3-3

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4-2.Quadratic Equations and Inequations

hard

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

$21$

$\frac{169}{8}$

$19$

$\frac{155}{8}$

(JEE MAIN-2023)

Solution

$\beta \gamma=1$

$\alpha=-1$

$\text { Put } \alpha=-1$

$-1-b+c=0$

$c-b=1$

$\text { also }$

$\alpha \cdot \beta \cdot \gamma=-c$

$-1=-c \Rightarrow c=1$

$\therefore b=0$

$x^3+1=0$

$\alpha=-1, \beta=- w , \gamma=- w ^2$

$\therefore b^3+2 c^3-3 \alpha^3-6 \beta^3-8 \gamma^3$

$0+2+3+6+8=19$

Standard 11

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Solution

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