If $\alpha,\ \beta,\ \gamma$ are the cube roots of a negative number $p$, then for any three real numbers $x,\ y,\ z$ the value of $\frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha}$ is

Question

Dr. Abubacker · Accepted Answer

(b)
As $p < 0$, therefore $p = - q$, where $q > 0$
$	herefore\ \ \ p^{1/3} = ( - q)^{1/3} = q^{1/3}( - 1)^{1/3}\$
$\ \  \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p^{1/3} = - q^{1/3},\  - q^{1/3}\omega,\  - q^{1/3}\omega^{2}$
Let $\alpha = - q^{1/3}$, $\beta =$ $q^{1/3}\omega,\ \$and$\ \ \gamma = - q^{1/3}\omega^{2}$
$	herefore\ \ \ \frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha} = \frac{x + y\omega + z\omega^{2}}{x\omega + y\omega^{2} + z} = \omega^{2} = \frac{- 1 - i\sqrt{3}}{2}$

Q. If $\alpha,\ \beta,\ \gamma$ are the cube roots of a negative number $p$, then for any three real numbers $x,\ y,\ z$ the value of $\frac{x\alpha + y\beta + z\gamma}{x\beta + y\gamma + z\alpha}$ is

Explanation

Reviewed by Dr. Abubacker

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