If omega is a complex cube root of unity,then | vedclass

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(B) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.We need to evaluate the expression: $(x

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$x^3+1$

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(B) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.We need to evaluate the expression: $(x+1)(x+\omega)(x-\omega-1)$.First,rewrite the third term: $(x - (\omega + 1))$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + 1 = -\omega^2$.So the expression becomes $(x+1)(x+\omega)(x - (-\omega^2)) = (x+1)(x+\omega)(x+\omega^2)$.Now,multiply $(x+\omega)(x+\omega^2) = x^2 + \omega^2 x + \omega x + \omega^3 = x^2 + x(\omega + \omega^2) + 1$.Since $\omega + \omega^2 = -1$ and $\omega^3 = 1$,this simplifies to $x^2 - x + 1$.Finally,multiply by $(x+1)$: $(x+1)(x^2 - x + 1) = x^3 + 1$.

If $\omega$ is a complex cube root of unity,then $(x+1)(x+\omega)(x-\omega-1)$ is equal to

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