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. $pi$

Answer 1

Answer: (B)

$\frac{-1-i\sqrt{3}}{2}$

Answer 2

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