If alpha is a non-real root of the equation x | vedclass

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

Question

(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The given equation is $x^6-1=0$,which implies $x^6=1$. Since $\alpha$ is a root,$\alpha^6=1$. Also,$\alpha 
eq 1$ because $\alpha$ is a non-real root (the roots of $x^6-1=0$ are $e^{i2k\pi/6}$ for $k=0, 1, 2, 3, 4, 5$,and only $k=0$ gives $x=1$). We can write the numerator as $\alpha^2(1+\alpha+\alpha^2+\alpha^3)$. Since $\alpha^6-1 = (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1) = 0$ and $\alpha 
eq 1$,we have $\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1 = 0$. This implies $\alpha^5+\alpha^4+\alpha^3+\alpha^2 = -\alpha-1 = -(\alpha+1)$. Substituting this into the expression: $\frac{-(\alpha+1)}{\alpha+1} = -1$.

If $\alpha$ is a non-real root of the equation $x^6-1=0$,then $\frac{\alpha^2+\alpha^3+\alpha^4+\alpha^5}{\alpha+1} = $

Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then $(2-\omega)^2(2-\omega^2)^2(2-\omega^{10})^2(2-\omega^{11})^2=$

If $\alpha, \beta$ are the roots of the equation $x^2-2x+2=0$,then $\alpha^{2020}+\beta^{2020}=$

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

If ${\left( {\frac{{1 + \cos \theta + i\sin \theta }}{{i + \sin \theta + i\cos \theta }}} \right)^4} = \cos n\theta + i\sin n\theta $,then $n$ is equal to

If $\alpha$ satisfies the equation $x^2+x+1=0$ and $(1+\alpha)^7=A+B\alpha+C\alpha^2$,where $A, B, C \geq 0$,then $5(3A-2B-C)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module