Let x = alpha + beta,y = alpha omega + beta o | vedclass

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Question

(C) Given $x = \alpha + \beta$,$y = \alpha \omega + \beta \omega^2$,and $z = \alpha \omega^2 + \beta \omega$. We know that $1 + \omega + \omega^2 = 0$

Vedclass · Accepted Answer

$\alpha^3 + \beta^3$

Vedclass · Answer

(C) Given $x = \alpha + \beta$,$y = \alpha \omega + \beta \omega^2$,and $z = \alpha \omega^2 + \beta \omega$. We know that $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. First,calculate the product $yz$: $yz = (\alpha \omega + \beta \omega^2)(\alpha \omega^2 + \beta \omega)$ $yz = \alpha^2 \omega^3 + \alpha \beta \omega^2 + \alpha \beta \omega^4 + \beta^2 \omega^3$ Since $\omega^3 = 1$ and $\omega^4 = \omega$,we get: $yz = \alpha^2(1) + \alpha \beta \omega^2 + \alpha \beta \omega + \beta^2(1)$ $yz = \alpha^2 + \alpha \beta(\omega^2 + \omega) + \beta^2$ Since $\omega^2 + \omega = -1$,we have: $yz = \alpha^2 - \alpha \beta + \beta^2$ Now,calculate $xyz = x(yz)$: $xyz = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$ Using the algebraic identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$,we get: $xyz = \alpha^3 + \beta^3$.

Let $x = \alpha + \beta$,$y = \alpha \omega + \beta \omega^2$,and $z = \alpha \omega^2 + \beta \omega$,where $\omega$ is an imaginary cube root of unity. The product $xyz$ is equal to:

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