If the cube root of 1 is omega,then the value | vedclass

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

Question

(C) Given expression: $(3 + \omega + 3\omega^2)^4$. We know that $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$. Also,$3 + 3\omega

Vedclass · Accepted Answer

$16\omega$

Vedclass · Answer

(C) Given expression: $(3 + \omega + 3\omega^2)^4$. We know that $1 + \omega + \omega^2 = 0$,which implies $\omega + \omega^2 = -1$. Also,$3 + 3\omega^2 = 3(1 + \omega^2) = 3(-\omega) = -3\omega$. Substituting this into the expression: $(3 + 3\omega^2 + \omega)^4 = (-3\omega + \omega)^4$. $= (-2\omega)^4$. $= 16\omega^4$. Since $\omega^3 = 1$,we have $\omega^4 = \omega^3 	imes \omega = 1 	imes \omega = \omega$. Therefore,the value is $16\omega$.

If the cube root of $1$ is $\omega$,then the value of $(3 + \omega + 3\omega^2)^4$ is

Similar Questions

If $x+iy=(-1+i\sqrt{3})^{2010}$,then $x$ is

If $\omega$ is a cube root of unity,then $(1 + \omega)^3 - (1 + \omega^2)^3 = $

The least positive integer $n$ for which $\left( \frac{1 + i\sqrt{3}}{1 - i\sqrt{3}} \right)^n = 1$ is?

The real part of $\frac{(\cos a+i \sin a)^6}{(\sin b+i \cos b)^8}$ is

$\sum_{k=1}^{6} (\sin \frac{2 \pi k}{7} - i \cos \frac{2 \pi k}{7}) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module