If x = a + b,y = aomega + bomega^2,and z = ao | vedclass

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(B) Given $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$. We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Expanding

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

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(B) Given $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$. We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. Expanding the cubes: $x^3 = (a + b)^3 = a^3 + b^3 + 3ab(a + b)$ $y^3 = (a\omega + b\omega^2)^3 = a^3\omega^3 + b^3\omega^6 + 3a^2b\omega^4 + 3ab^2\omega^5 = a^3 + b^3 + 3a^2b\omega + 3ab^2\omega^2$ $z^3 = (a\omega^2 + b\omega)^3 = a^3\omega^6 + b^3\omega^3 + 3a^2b\omega^5 + 3ab^2\omega^4 = a^3 + b^3 + 3a^2b\omega^2 + 3ab^2\omega$ Adding these: $x^3 + y^3 + z^3 = 3(a^3 + b^3) + 3a^2b(1 + \omega + \omega^2) + 3ab^2(1 + \omega^2 + \omega)$ Since $1 + \omega + \omega^2 = 0$,the terms with $a^2b$ and $ab^2$ vanish. Thus,$x^3 + y^3 + z^3 = 3(a^3 + b^3)$.

If $x = a + b$,$y = a\omega + b\omega^2$,and $z = a\omega^2 + b\omega$,then the value of $x^3 + y^3 + z^3$ is equal to

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