First slide

Theory of equations

Question

$If α, β, γ are roots of the cubic x^{3} - 2 x + 3 = 0, then the value of \frac{1}{α^{3} + β^{3} + 6} + \frac{1}{β^{3} + γ^{3} + 6} + \frac{1}{γ^{3} + α^{3} + 6} equals to$

Moderate

A

$\frac{1}{3}$
B

$- \frac{1}{3}$
C

$\frac{1}{2}$
D

$- \frac{1}{2}$

Solution

$\begin{array}{l} Since α, β, γ are the roots of x^{3} - 2 x + 3 = 0 \dots \dots (1) \\ Then S_{1} = α + β + γ = 0, S_{2} = α β + β γ + γ α = - 2, S_{3} = α β γ = - 3 \end{array}$
$Substituting α, β in equation (1), we get α^{3} = 2 α - 3, β^{3} = 2 β - 3$
$\begin{array}{l} ∴ α^{3} + β^{3} = 2 (α + β) - 6 \\ \Rightarrow α^{3} + β^{3} + 6 = 2 (α + β) = 2 (- γ) \\ ∴ \sum \frac{1}{α^{3} + β^{3} + 6} = \sum - \frac{1}{2 γ} \\ = - \frac{1}{2} [\frac{1}{α} + \frac{1}{β} + \frac{1}{γ}] \\ = - \frac{1}{2} [\frac{α β + β γ + γ α}{α β γ}] \\ = - \frac{1}{2} [\frac{- 2}{- 3}] \\ = - \frac{1}{3} \end{array}$

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