If 1, omega, omega^2 are the cube roots of un | vedclass

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(A) Given $\alpha = \omega + 2\omega^2 - 3$. Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$. Substituting this into the expression

Vedclass · Accepted Answer

-$63$

Vedclass · Answer

(A) Given $\alpha = \omega + 2\omega^2 - 3$. Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$. Substituting this into the expression for $\alpha$: $\alpha = (\omega + \omega^2) + \omega^2 - 3 = -1 + \omega^2 - 3 = \omega^2 - 4$. Thus,$\omega^2 = \alpha + 4$. Cubing both sides,we get $(\omega^2)^3 = (\alpha + 4)^3$. Since $\omega^3 = 1$,we have $\omega^6 = 1$. So,$1 = \alpha^3 + 3(\alpha^2)(4) + 3(\alpha)(4^2) + 4^3$. $1 = \alpha^3 + 12\alpha^2 + 48\alpha + 64$. Subtracting $61$ from both sides: $\alpha^3 + 12\alpha^2 + 48\alpha + 3 = 1 - 64 = -63$.

If $1, \omega, \omega^2$ are the cube roots of unity and $\alpha = \omega + 2\omega^2 - 3$,then $\alpha^3 + 12\alpha^2 + 48\alpha + 3$ equals

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