Question

If $\alpha, \,\beta, \,\gamma$ are the cube roots of a positive number $p$, then for any real $x,\, y,\, z$ the expression $\left(\frac{\alpha x+\beta y+\gamma z}{\beta x+\gamma y+\alpha z}\right)$ equals:

• A. $\frac{-1-\sqrt{3} i}{2}$
• B. $\frac{-1+\sqrt{3} i}{2}$
• C. $\frac{1+\sqrt{3} i}{2}$
• D. $\frac{1-\sqrt{3} i}{2}$

Answer 1

Answer: (A)

Given $\alpha,\, \beta,\, \gamma$ are the cube roots of a positive number $p$ and $x, y, z$ are real numbers.
Since $\alpha, \,\beta,\, \gamma$ be the cube roots of a positive number $p$.
$\therefore \, \alpha=p^{1 / 3}, \,\beta=\omega p^{1 / 3}, \gamma=\omega^{2} p^{1 / 3}$
So, $\frac{\alpha x+\beta y+\gamma z}{\beta x+\gamma y+\alpha \,z}$
$=\frac{p^{1 / 3} x+\omega p^{1 / 3} y+\omega^{2} p^{1 / 3} z}{\omega p^{1 / 3} x+\omega^{2} p^{1 / 3} y+p^{1 / 3} z}$
$=\frac{x+\omega y+\omega^{2} z}{\omega x+\omega^{2} y+z}$
$=\frac{\omega\left(x+\omega y+\omega^{2} z\right)}{\omega\left(\omega x+\omega^{2} y+z\right)}=\frac{1}{\omega}=\frac{\omega^{2}}{\omega^{3}}=\omega^{2}$
$=\frac{-1-i \sqrt{3}}{2}$