α, β, γ are the roots of the equation x^3+2 x^2-x-2=0 then α^6+β^6+γ^6=

Correct answer: (C) 68. α+β+γ=-2 ; α β+β γ+γ α=-1 ; α β γ=2 x^3+2 x^2-x-2=0 ⇒(x+1)(x-1)(x+2)=0 ∴ x=1,-1,-2 α^6+β^6+γ^6=(1)^6+(-1)^6+(-2)^6=66

Q57 / 160 · TS EAMCET 2023 (14 May Shift 1) · Mathematics

$\alpha, \beta, \gamma$ are the roots of the equation $x^3+2 x^2-x-2=0$ then $\alpha^6+\beta^6+\gamma^6=$

A
$3$
B
$129$
Correct answer
C
$68$
D
$192$

Solution

$\alpha+\beta+\gamma=-2 ; \alpha \beta+\beta \gamma+\gamma \alpha=-1 ; \alpha \beta \gamma=2$ $\begin{aligned} & x^3+2 x^2-x-2=0 \\ & \Rightarrow(x+1)(x-1)(x+2)=0 \\ & \therefore x=1,-1,-2 \\ & \alpha^6+\beta^6+\gamma^6=(1)^6+(-1)^6+(-2)^6=66\end{aligned}$

Hence, the correct answer is option: 'C'.

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