Change of coordinates in a polynomial ring I want to prove that $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x\in \mathbb{C}[x,y,z]$$ can be transformed into $$y^2z+x^3+\alpha xz^2+\beta z^3$$ by means of an invertible linear change of coordinates.
I tried to complete the square in $y$, I got $z(y^2+yz+xy)=z(y+t)^2-zt^2$ where $t=(z+x)/2$, but I don't know how to proceed.
I also tried to substitute $x+py+qz$ for $x$ and expand. In the resulting polynomial, I got the terms $3p^2xy^2$ and $3px^2y$ which must be zero, so $p=0$. But then there are also the terms $(d+3q)x^2z$ and $(bq+1)yz^2$ which also must vanish simultaneously, which is impossible. Also I got the term $bxyz$ which must vanish, but $b$ is not necessarily zero...
 A: Rewrite $$zy^2+z^2y+bxyz+x^3+cz^3+dx^2z+az^2x=0$$ as $$y^2z+bxyz+yz^2=x^3+dx^2z+axz^2+cz^3.$$ Make the change of coordinates $y$ goes to $y-\frac{bx+z}{2}$. This gives $$y^2z-\frac{bxz^2}{2}-\frac{b^2x^2z}{4}-\frac{z^3}{4} = x^3+dx^2z+axz^2+cz^3$$
Or equivalently, after collecting terms, $$y^2z=x^3+\left(d+\frac{b^2}{4}\right)x^2z+\left(a+\frac{b}{2}\right)xz^2+\frac{4c+1}{4}z^3$$
Setting $p=d+\frac{b^2}{4}$, $q=a+\frac{b}{2}$, and $r=c+\frac{1}{4}$, we may replace $x$ by $x-\frac{p}{3}$ to get an equation of the form
$$y^2z=x^3+\alpha xz^2+\beta z^3$$
Both coordinate changes here are instances of the following general fact: If $p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$, you can control the coefficient $a_{n-1}$ by linear changes of variables. Recall that $a_{n-1}$ is the negative of the sum of the roots: if we make the substitution $x\mapsto x-\frac{a_{n-1}}{n}$, the sum of the new roots is therefore $n\cdot\frac{-a_{n-1}}{n}=-a_{n-1}$ more than the old sum, which means it is zero.
The application of this to the RHS should be somewhat obvious. On the left, note that we're really applying this to $y^2+(bxz+z)y$, as $y^2z+bxyz+yz^2=z(y^2+(bxz+z)y)$.
