Take a curve $C$ (i.e. an irreducible algebraic set) in $\mathbb{P}(k)^2$ (projective plane of some algebraically closed field $k$). For $F\in k[X,Y,Z]$ denote by $V(F)$ its zeroes in $\mathbb{P}^2$, and for $P\in V(F)$ denote $m_P(F)$ its multiplicity on $F$. Let $$P=[0:0:1],\hspace{0.1cm} P'=[0:1:0],\hspace{0.1cm} P''=[1:0:0],$$ $$L=V(Z),\hspace{0.1cm} L'=V(Y),\hspace{0.1cm} L''=V(X),$$ $$r=m_{P}(C),\hspace{0.1cm} r'=m_{P'}(C), \hspace{0.1cm}r''=m_{P''}(C).$$ Define $Q:\mathbb{P}^2\setminus\{P,P',P''\}\longrightarrow \mathbb{P}^2$ by $[x:y:z]\mapsto[yz:xz:xy]$.
Suppose $F$ is an irreducible form of degree $n$ such that $C=V(F)$, then $F^Q\stackrel{\text{def}}{=}F\circ Q$ is a form of degree 2n. Additionaly: $$F^Q=Z^rY^{r'}X^{r''}F'$$ where $X,Y$ and $Z$ do not divide $F'$ (needs a proof but it's not important for my question). This $F'$ is called the proper transform of $F$.
My question is: It is stated (page 88 item 3) that $$F'=\sum_{i=0}^{n-r}=F_{r+i}(Y,X)X^{n-r-r''-i}Y^{n-r-r'}Z^{i}.$$ where $F_j\in k[X,Y]$ is a form of degree $j$. I can't prove that $F'$ can be written in this way, and it seems to me to contradict the fact that $X$ and $Y$ do not divide $F'$, as $r''<n$ is enough to imply that $X$ divides that sum, for example.
I'd appreciate any help!
As explained at page 88, you write $$F(X,Y,Z)=F_r(X,Y)Z^{n-r}+F_{r+1}(X,Y)Z^{n-r-1}+\cdots+F_n(X,Y)=\sum\limits_{i=0}^{n-r} F_{r+i}(X,Y)Z^{n-r-i}$$ where $F_i$ is homogeneous of degree $i$. Moreover, $F_r\not=0$ as $r=m_p(C)$.
This gives $F^Q=F(YZ,XZ,XY)=\sum\limits_{i=0}^{n-r} F_{r+i}(YZ,XZ)(XY)^{d-r-i}=\sum\limits_{i=0}^{n-r} Z^{r+i}F_{r+i}(Y,X)(XY)^{d-r-i}$
As $F^Q=Z^rY^{r'}X^{r''}F'$ you obtain
$$F'=\sum\limits_{i=0}^{n-r} F_{r+i}(Y,X)X^{d-r-i-r'}Y^{d-r-i-r''}Z^{i}$$ as desired.
By construction, $X$ does not divide $F'$ and $Y$ does not divide it either.