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

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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

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