Intersection number between two projective curves I am trying to find all the intersection points (over $\mathbb{C}$) with the corresponding intersection numbers between the curves:
$F: (X^2 + Y^2)Z + X^3 + Y^3$
and
$G: X^3 + Y^3 - 2XYZ$
Well, the way I am doing this is:
Let's consider the case $Z=0$ (points at infinity):
Then the two curves turn into $X^3 + Y^3=0$ and then we have three points $[1:-1:0]$,$[1:c:0]$,$[1:c^2:0]$, c a primitive 3rd root of unity. 
The case $Z=1$:
We then get:
$F: (X^2 + Y^2) + X^3 + Y^3$
and
$G: X^3 + Y^3 - 2XY$
And it is easy to see that the only point in the intersection is $[0:0:1]$.
Computing $I([0:0:1],F\cap G)$, which is the same as $I((0,0), f\cap g)$, where $g = G(x,y,1)$ and $f = F(x,y,1)$:
$I((0,0), f\cap g) = I((0,0), g\cap (f-g) ) = I((0,0), g \cap (X+Y)^2 ) = 2I((0,0), g\cap (X+Y))$, and since $g = X^3 + Y^3 - 2XY$, its tangent lines at $(0,0)$ are the $x$ and $y$ axis whereas $X+Y$ is its own tangent line at $(0,0)$, so since there is no tangent in common, $I((0,0), g\cap (X+Y)) = m_{(0,0)}(g)=2$, thus $I((0,0), f\cap g) = 4$. 
Now how do I proceed to compute the intersection number of the remaining points? I tried to dehomogenize the polynomials with respect to $X$ because the points are of the form $[1:x:0]$, but I can't go further since the ways I see to find this number would give me the same number to all of the three remaining points, and this is impossible since the sum of the intersection points is 9 by Bezout, but the point $[0:0:1]$ count with 4 on this calculation, remaining 5 to ditribute between the three remaining points. Can anyone help me?
 A: Let $P_1 = [1:-1:0], P_2 = [1:\zeta:0], P_3 = [1:\zeta^{-1}:0]$, where $\zeta$ is a primitive $6^\text{th}$ root of unity.  (Note that you have a small error in your points of intersection: the second coordinates should be $-\omega$ and $-\omega^2$ where $\omega$ is a primitive $3^\text{rd}$ root of unity.  Since $-\omega^2 = \zeta$, this is the same as what I have above.)  I claim that $I_{P_1}(F,G) = 3$ and $I_{P_2}(F,G) = I_{P_3}(F,G) = 1$, so the intersection numbers sum to $4 + 3 + 1 + 1 = 9$, as we would hope.
Since $F-G = (X+Y)^2 Z =: H$, we can instead find the intersection multiplicities for $G$ and $H$.  We work in the affine open subset where $X \neq 0$ with coordinates $u = Y/X$ and $v = Z/X$.  Dehomogenizing $G$ and $H$, we have
$$
g = u^3 - 2uv + 1 \qquad h = (1+u)^2 v \, .
$$
Let's look at $P_1$ first. We make the change of coordinate $u \leftarrow u+1$ so that the origin is now the point of interest.  Under this change of coordinate, we have
$$
g = (u-1)^3 - 2(u-1)v + 1 = u^3 - 3u^2 - 2uv + 3u + 2v
$$
and $h = u^2v$.  Then
$$
I_{P_1}(g,h) = 2I_{P_1}(g,u) + I_{P_1}(g,v) = 2 + 1 = 3
$$
since $g$ has no common tangents with $u$ or $v$.
Let's consider $P_2$ now.  We again make the change of coordinate $u \leftarrow u- \zeta$ to work at the origin.  Then
\begin{align*}
g &= (u+\zeta)^3 - 2(u+\zeta)v + 1 = u^3 + 3 \zeta u^2 - 2uv + 3 \zeta^2 u - 2 \zeta v\\
h &= (u + \zeta)^2 v + 2(u + \zeta)v + v = u^2 v + 2(\zeta + 1) uv + (\zeta + 1)^2 v = u^2 v + 2(\zeta + 1) uv + 3 \zeta v
\end{align*}
since
$$
(\zeta + 1)^2 = \zeta^2 + 2 \zeta + 1 = \overbrace{\zeta^2 - \zeta + 1}^0 + 3 \zeta \, .
$$
From these expressions for $g$ and $h$ we see that $P_2$ has multiplicity $1$ for both $g$ and $h$.  Since $g$ and $h$ have no common tangents, then $I_{P_2}(g,h) = 1$.
I haven't checked $P_3$, but I think it should be the same as for $P_2$: both $\zeta$ and $\zeta^{-1}$ are primitive $6^\text{th}$ roots of unity, so they should be algebraically indistinguishable.
