Let $p,q,r$ coprime polynomials. **Show that $p^3-q^2=r^6$ violate Mason-Stothers Theorem.
My attempt:
We know we are in the condictions of the Mason's Theorem. Then, $\max\{3\deg p,2\deg r,6\deg d\}\leq N_0(p^3r^2d^6)−1$. Since $N_0(f)$ is the number of distinct roots of $f$, we will have that $N_0(p^3r^2d^6)=N_0(pr^2d^2)\leq\deg p+2\deg r+2\deg d$ and then
$$3\deg p+2\deg r+6\deg r\leq3\deg p+3\deg r+6\deg r−3⇒3≤degr.$$
Am I in the right path to find the contradiction?
It seems like this is easier than you are making it. Let $P=\deg p, Q=\deg q, R= \deg r.$ Then we know $$\max(3P, 2Q, 6R) \le N_0(p^3q^2r^6)-1=N_0(pqr)-1 =P+Q+R-1\tag 1$$
Now suppose $\max(3P, 2Q, 6R)=3P.$ Then we have $$ \begin{align} &2Q\le 3P \implies Q\le \frac{3P}{2}\\ &6R\le 3P \implies R \le \frac{P}{2} \end{align}$$ and substituting into (1) gives $$3P<=P+\frac{3P}{2}+\frac{P}{2}-1=3P-1,$$ contradiction. We deduce similar contradictions on assuming that the maximum is $2Q$ or $6R.$