(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 $

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(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 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

Medium

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
$192$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

4-2.Quadratic Equations and Inequations

Subtopic

Solution of quadratic equations and Nature of roots

Previous Year

TS EAMCET 2023 Paper All TS EAMCET years →

$\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 3 x^2 - 10 x - 24 = 0$. If $\alpha > \beta > \gamma$ and $\alpha^3 + 3 \beta^2 - 10 \gamma - 24 = 11 k$,then $k = $

EasyTS EAMCET 2024

View Solution

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

DifficultTS EAMCET 2009

View Solution

The number of solutions for the equation $x^2-5|x|+6=0$ is .........

EasyAP EAMCET 2020

View Solution

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

DifficultAIEEE 2007

View Solution

If both roots of the equation $x^2 + \lambda x + \mu = 0$ are equal and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = \dots$

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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

Similar Questions

$\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 3 x^2 - 10 x - 24 = 0$. If $\alpha > \beta > \gamma$ and $\alpha^3 + 3 \beta^2 - 10 \gamma - 24 = 11 k$,then $k = $

Let $f(x) = x^2 + ax + b$,where $a, b \in R$. If $f(x) = 0$ has all its roots imaginary,then the roots of $f(x) + f'(x) + f''(x) = 0$ are

The number of solutions for the equation $x^2-5|x|+6=0$ is .........

If the difference between the roots of the equation $x^2 + ax + 1 = 0$ is less than $\sqrt{5}$,then the set of possible values of $a$ is

If both roots of the equation $x^2 + \lambda x + \mu = 0$ are equal and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module