If omega is an imaginary cube root of unity,( | vedclass

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

Question

(D) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.Substituting this into the expression:$(1 + \omega - \omega^2)^7 =

Vedclass · Accepted Answer

$-128\omega^2$

Vedclass · Answer

(D) We know that $1 + \omega + \omega^2 = 0$,which implies $1 + \omega = -\omega^2$.Substituting this into the expression:$(1 + \omega - \omega^2)^7 = (-\omega^2 - \omega^2)^7$$= (-2\omega^2)^7$$= (-2)^7 	imes (\omega^2)^7$$= -128 	imes \omega^{14}$Since $\omega^3 = 1$,we have $\omega^{14} = (\omega^3)^4 	imes \omega^2 = 1^4 	imes \omega^2 = \omega^2$.Therefore,the expression equals $-128\omega^2$.

If $\omega$ is an imaginary cube root of unity,$(1 + \omega - \omega^2)^7$ equals

Similar Questions

If $1, \omega, \omega^{2}$ are the cube roots of unity,then $(1+\omega)(1+\omega^{2})(1+\omega^{4})(1+\omega^{8})$ is equal to

For $n \in Z^{+}$,$(1+\sin \theta+i \cos \theta)^n+(1+\sin \theta-i \cos \theta)^n=$

If $z$ is a non-real root of $x^7=1$,then $1+3z+5z^2+7z^3+9z^4+11z^5+13z^6=$

$(\sqrt{\sqrt{2}+1} + i\sqrt{\sqrt{2}-1})^8 =$

$(-1 + i\sqrt{3})^{20}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module