(C) Given the cubic equation $x^3+2x^2-x-2=0$. Factorizing the equation: $x^2(x+2)-1(x+2)=0$ $(x^2-1)(x+2)=0$ $(x-1)(x+1)(x+2)=0$ Thus,the roots are $

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(C) Given the cubic equation $x^3+2x^2-x-2=0$. Factorizing the equation: $x^2(x+2)-1(x+2)=0$ $(x^2-1)(x+2)=0$ $(x-1)(x+1)(x+2)=0$ Thus,the roots are $\alpha=1, \beta=-1, \gamma=-2$. We need to calculate $\alpha^6+\beta^6+\gamma^6$: $\alpha^6+\beta^6+\gamma^6 = (1)^6 + (-1)^6 + (-2)^6$ $= 1 + 1 + 64$ $= 66$

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Solution of quadratic equations and Nature of roots

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

TS EAMCET 2023 All TS EAMCET

A
$3$
B
$129$
C
$66$
D
$192$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Solution of quadratic equations and Nature of roots

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

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