If alpha, beta, gamma are the cube roots of p | vedclass

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(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega =

Vedclass · Accepted Answer

None of these

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(D) Let $p = -q$ where $q > 0$. The cube roots of $p$ are $\alpha = -q^{1/3}$,$\beta = -q^{1/3}\omega$,and $\gamma = -q^{1/3}\omega^2$,where $\omega = \frac{-1 + i\sqrt{3}}{2}$ is the complex cube root of unity.Substituting these into the expression:$\frac{x(-q^{1/3}) + y(-q^{1/3}\omega) + z(-q^{1/3}\omega^2)}{x(-q^{1/3}\omega) + y(-q^{1/3}\omega^2) + z(-q^{1/3}\omega^3)} = \frac{-(x + y\omega + z\omega^2)}{-\omega(x + y\omega + z\omega^2)} = \frac{1}{\omega} = \omega^2$.Since $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$,none of the given options match.

If $\alpha, \beta, \gamma$ are the cube roots of $p$ $(p < 0)$,then for any $x, y$ and $z$,$\frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = $

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