Let 1, omega and omega^2 be the cube roots of | vedclass

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(C) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$,which implies $\omega^{-1}=\omega

Vedclass · Accepted Answer

$-64 \omega^{-1}$

Vedclass · Answer

(C) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2=0$ and $\omega^3=1$,which implies $\omega^{-1}=\omega^2$.Substituting $\omega^{-1}=\omega^2$ into the expression:$(1-\omega+\omega^2)^5-2(1+\omega-\omega^2)^4$Using $1+\omega^2=-\omega$ and $1+\omega=-\omega^2$:$(-\omega-\omega)^5-2(-\omega^2-\omega^2)^4$$=(-2\omega)^5-2(-2\omega^2)^4$$=-32\omega^5-2(16\omega^8)$$=-32\omega^2-32\omega^2$$=-64\omega^2$Since $\omega^2=\omega^{-1}$,the result is $-64\omega^{-1}$.

Let $1, \omega$ and $\omega^2$ be the cube roots of unity. What is the value of $(1-\omega+\omega^{-1})^5-2(1+\omega-\omega^{-1})^4$?

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