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, $\frac{\alpha x+\beta y+\gamma z}{\beta x+\gamma y+\alpha z}$
$=\frac{p^{1/3}x+\omega p^{1/3}y+{\omega }^2{p}^{1/3}z}{\omega p^{1/3}x+{\omega }^2{p}^{1/3}y+p^{1/3}z}$
$=\frac{x+\omega y+{\omega }^{2}z}{\omega x+{\omega }^{2}y+z}$
$=\frac{\omega \left(x+\omega y+{\omega }^{2}z\right)}{\omega \left(\omega x+{\omega }^{2}y+z\right)}=\frac{1}{\omega }=\frac{{\omega }^2}{{\omega }^3}={\omega }^2$
$=\frac{-1-i\sqrt{3}}{2}$