If $\alpha , \beta , \gamma$are the cube roots of $p, p < 0,$then for any x, y and z which does not make denominator zero,

 the expression  $\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }$

$x^3=p={\left(p^{\frac{1}{3}}\right)}^3\Rightarrow x=p^{\frac{1}{3}},p^{\frac{1}{3}}\omega ,p^{\frac{1}{3}}{\omega }^2$

Let $\alpha =p^{\frac{1}{3}},\beta =p^{\frac{1}{3}}\omega ,\gamma =p^{\frac{1}{3}}{\omega }^2$ 

$\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }=\frac{x+y\omega +{\omega }^{2}z}{x\omega +y{\omega }^2+z}=\frac{1}{\omega }={\omega }^2$

If $\alpha =p^{\frac{1}{3}}\omega ,\beta =p^{\frac{1}{3}},\gamma =p^{\frac{1}{3}}{\omega }^2$ then , 

$\frac{x\alpha +y\beta +z\gamma }{x\beta +y\gamma +z\alpha }=\frac{1}{{\omega }^2}=\omega .$