Question

If $\alpha, \beta, \gamma$ are the cube roots of $p, p<0$, then for any $x, y$ and $z$ which does not make denominator zero, the expression $\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}$ equals

• A. $\omega, 1$
• B. $\omega, \omega^2$
• C. $\omega^2, 1$
• D. $1, \omega, \omega^2$

Answer 1

Answer: (B)

$x^3=p=\left(p^{\frac{1}{3}}\right)^3 \Rightarrow x=p^{\frac{1}{3}}, p^{\frac{1}{3}} \omega, p^{\frac{1}{3}} \omega^2$
Let $\alpha=p^{\frac{1}{3}}, \beta=p^{\frac{1}{3}} \omega, \gamma=p^{\frac{1}{3}} \omega^2$.
$\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}=\frac{x+y \omega+\omega^2 z}{x \omega+y \omega^2+z}=\frac{1}{\omega}=\omega^2$
If $\alpha=p^{\frac{1}{3}} \omega, \beta=p^{\frac{1}{3}}, \gamma=p^{\frac{1}{3}} \omega^2$, then $\frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}=\frac{1}{\omega^2}=\omega$