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} $