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(B) Given $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ where $\alpha=\omega$ and $\beta=\omega^2$ are the complex cube roots of unity. Since $1+\

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$3(a^3+b^3)$

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(B) Given $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ where $\alpha=\omega$ and $\beta=\omega^2$ are the complex cube roots of unity. Since $1+\omega+\omega^2=0$,we have $x+y+z = (a+b) + (a\omega+b\omega^2) + (a\omega^2+b\omega) = a(1+\omega+\omega^2) + b(1+\omega+\omega^2) = 0$. We know that if $x+y+z=0$,then $x^3+y^3+z^3=3xyz$. $3xyz = 3(a+b)(a\omega+b\omega^2)(a\omega^2+b\omega)$ $= 3(a+b)(a^2\omega^3 + ab\omega^2 + ab\omega^4 + b^2\omega^3)$ $= 3(a+b)(a^2 + ab\omega^2 + ab\omega + b^2)$ $= 3(a+b)(a^2 + ab(\omega^2+\omega) + b^2)$ Since $\omega^2+\omega = -1$,$= 3(a+b)(a^2 - ab + b^2) = 3(a^3+b^3)$.

If $x=a+b$,$y=a \alpha+b \beta$,$z=a \beta+b \alpha$ and $\alpha, \beta$ are the complex cube roots of unity,then $x^3+y^3+z^3=$

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