What is $\frac{x-1}{x^{q/p}-1}$ equal to? Given two positive integers $p, q$ with $p>q$ and $(p,q)=1$. Can we write $\frac{x-1}{x^{q/p}-1}$ into some thing like $x^{1-q/p}+x^{1-2q/p}+\cdots+1$? I guess the answer is no, but is there a polynomial  expression (fractional power is allowed) for $\frac{x-1}{x^{q/p}-1}$?
 A: The question is not very clear, so I am assuming mean this:
Is there a polynomial $P(x)$ (of finite degree, in $\mathbb{C}[x]$) such that
$$(x-1) = (x^{q/p} - 1) P(x^r)$$
for some rational $r$?
The answer is NO.
set $x = z^p$ and we get
$$(z^p - 1) = (z^q - 1)P(z^{rp})$$
Choose a (complex $z$) such that $z^q = 1$ and $z^p \neq 1$.
This is possible if $(p,q) = 1$ and $q \neq 1$.
A: Well, there's this,
$$\rm \frac{x-1}{x^{q/p}-1}=\frac{x-1}{x^{1/p}-1} \, \frac{x^{1/p}-1}{x^{q/p}-1}=\frac{1+x^{1/p}+\cdots+x^{p/p}}{1+x^{1/p}+\cdots+x^{q/p}}.$$
Or, formally
$$\rm (1-x^{\color{Red}1})\big(1-x^{\color{LimeGreen}1\color{Blue}{q/p}}+x^{\color{LimeGreen}2\color{Blue}{q/p}}-x^{\color{LimeGreen}3\color{Blue}{q/p}}+\cdots\big),$$
which doesn't cancel out into a polynomial (in fractional powers) because $\rm \color{Red}1+\color{LimeGreen}m\,\color{Blue}{q/p}=\color{LimeGreen}{n}\color{Blue}{q/p}$ has no solution in $\rm m,n$ (to see why reduce it modulo $\rm q$). The above of course has convergence restrictions.
