If $\alpha ,\beta ,\gamma$ are the roots of $x^3-3x^2+3x+7=0$ and $\omega$ is a complex cube root of unity, then the value of $(\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1})$ is equal to $p\left({\omega }^q\right)$ for some $p,q\in Z.$ Then the number of ordered pairs $(p,q)$ such that $p+q=15$ is

$\alpha ,\beta ,\gamma$ are roots of ${(x-1)}^3=-8$

$\Rightarrow \alpha -1,\beta -1,\gamma -1$ are equal to $-2,-2\omega ,-2{\omega }^2$ in any order then $\frac{\alpha -1}{\beta -1}+\frac{\beta -1}{\gamma -1}+\frac{\gamma -1}{\alpha -1}=3{\omega }^{2}or3\omega =3{\left(\omega \right)}^{3k+1}or3{\left(\omega \right)}^{3k+2}wherek\in Z$

$\therefore p=3,q=3k+1or3k+2$

$p+q=15\Rightarrow k=\frac{11}{3}or\frac{10}{3}$ which is not possible