If x = a + b,y = aalpha + bbeta,and z = abeta | vedclass

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(B) Given that $\alpha$ and $\beta$ are complex cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $1 + \omega

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$a^3 + b^3$

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(B) Given that $\alpha$ and $\beta$ are complex cube roots of unity,we have $\alpha = \omega$ and $\beta = \omega^2$ (or vice versa),where $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$.We need to calculate $xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega)$.First,multiply $y$ and $z$:$yz = (a\omega + b\omega^2)(a\omega^2 + b\omega) = a^2\omega^3 + ab\omega^2 + ab\omega^4 + b^2\omega^3$.Since $\omega^3 = 1$ and $\omega^4 = \omega$,this simplifies to:$yz = a^2(1) + ab(\omega^2 + \omega) + b^2(1) = a^2 + ab(-1) + b^2 = a^2 - ab + b^2$.Now,multiply by $x$:$xyz = (a + b)(a^2 - ab + b^2) = a^3 + b^3$.

If $x = a + b$,$y = a\alpha + b\beta$,and $z = a\beta + b\alpha$,where $\alpha$ and $\beta$ are complex cube roots of unity,then $xyz$ =

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