Question

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

• A. $\frac{1-i\sqrt{3}}{2}$
• B. $\frac{-1-i\sqrt{3}}{2}$
• C. $(x+y+z)i$
• D. $pi$
• E. $\frac{x+y+z}{2}pi$

Answer 1

Answer: (B)

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