solve $ax^3+by^3+cz^3+dx^2y+ex^2z+fxy^2+gxz^2+hy^2z+iyz^2=0$ for all triplets $(x,y,z)$. let $x,y,z$ be any 3 positive integers. If for all $x,y,z$, we have :
$$ax^3+by^3+cz^3+dx^2y+ex^2z+fxy^2+gxz^2+hy^2z+iyz^2=0$$ What can be said about the integral coefficients $a,b,c,e,f,g,h,i$? 
I think they must be all zeros. What do you think?
 A: Consider taking partial finite differences: for instance, evaluating at $z$ and $z+1$ and subtracting, it immediately follows that $c(3z^2+3z+1)+ex^2+(gx+iy)(2z+1)+hy^2=0$ for all $x,y,z$.  Just like with differentiation, the finite differencing operation lowers the degree of the equation; this equation is of degree $2$ (where the original was cubic).  Now you can take two finite differences in $y$ to immediately show that $h=0$ and similarly take two finite differences in $x$ to show that $e=0$, or take another two finite differences in $z$ to show that $c=0$.  In general, each possible third-order partial finite difference will annihilate all but one of your coefficients and allow you to show that that coefficient must be zero.
A: Using matrix language
$$\begin{pmatrix}a&b&c&d&e&f&g&h&i\end{pmatrix}\cdot \begin{pmatrix}x^4\\y^3\\z^3\\x^2y\\x^2z\\xy^2\\xz^2\\y^2z\\yz^2\end{pmatrix}=(0)$$
It is clear that for arbitrary values of $x,y,z$ we can get much more than nine linear equations for the nine coefficients $a,b,c,d,e,f,g,h,i$ so that the only possible compatibility be with $a=b=c=d=e=f=g=h=i=0$
A: A special case of the parametrization:
$${z}^{3}-2\,a\,y\,{z}^{2}-b\,x\,{z}^{2}+\left( {a}^{2}-b\right) \,{y}^{2}\,z+\left( a\,b-3\,c\right) \,x\,y\,z+a\,c\,{x}^{2}\,z+\left( c+a\,b\right) \,{y}^{3}+\left( a\,c+{b}^{2}\right) \,x\,{y}^{2}+2\,b\,c\,{x}^{2}\,y+{c}^{2}\,{x}^{3}=c\cdot\,\left( {x}_{1}^{3}\,{c}^{2}+2\,{x}_{1}^{2}\,{y}_{1}\,b\,c+{x}_{1}^{2}\,{z}_{1}\,a\,c+{x}_{1}\,{y}_{1}^{2}\,a\,c-3\,{x}_{1}\,{y}_{1}\,{z}_{1}\,c+{y}_{1}^{3}\,c+{x}_{1}\,{y}_{1}^{2}\,{b}^{2}+{x}_{1}\,{y}_{1}\,{z}_{1}\,a\,b+{y}_{1}^{3}\,a\,b-{x}_{1}\,{z}_{1}^{2}\,b-{y}_{1}^{2}\,{z}_{1}\,b+{y}_{1}^{2}\,{z}_{1}\,{a}^{2}-2\,{y}_{1}\,{z}_{1}^{2}\,a+{z}_{1}^{3}\right)\cdot \,\left( {x}_{2}^{3}\,{c}^{2}+2\,{x}_{2}^{2}\,{y}_{2}\,b\,c+{x}_{2}^{2}\,{z}_{2}\,a\,c+{x}_{2}\,{y}_{2}^{2}\,a\,c-3\,{x}_{2}\,{y}_{2}\,{z}_{2}\,c+{y}_{2}^{3}\,c+{x}_{2}\,{y}_{2}^{2}\,{b}^{2}+{x}_{2}\,{y}_{2}\,{z}_{2}\,a\,b+{y}_{2}^{3}\,a\,b-{x}_{2}\,{z}_{2}^{2}\,b-{y}_{2}^{2}\,{z}_{2}\,b+{y}_{2}^{2}\,{z}_{2}\,{a}^{2}-2\,{y}_{2}\,{z}_{2}^{2}\,a+{z}_{2}^{3}\right)$$
$$x={x}_{1}\,{x}_{2}\,c+{y}_{2}\,\left( {z}_{1}-{y}_{1}\,a\right) +{y}_{1}\,{z}_{2},$$
$$y={y}_{2}\,\left( {x}_{1}\,c+{y}_{1}\,b\right) +{y}_{1}\,{x}_{2}\,c+{z}_{1}\,{z}_{2},$$
$$z={y}_{2}\,\left( {y}_{1}\,c+{z}_{1}\,b\right) +{z}_{2}\,\left( {x}_{1}\,c+{y}_{1}\,b+{z}_{1}\,a\right) +{z}_{1}\,{x}_{2}\,c.$$
