multiplicity of the intersections between $p=x_0x_1^2+x_1x_2^2+x_2x_0^2\,\,$ and $\,\,q=-8(x_0^3+x_1^3+x_2^3)+24x_0x_1x_2$ in $\mathbb{P^2(K)}$ in $\mathbb{P^2(K)}$ where $\mathbb{K}$ is an algebraically closed field and $[x_0,x_1,x_2]$ the homogeneous coordinates, consider the following (homogeneous) polynomials:
$p=x_0x_1^2+x_1x_2^2+x_2x_0^2\,\,$ and $\,\,q=-8(x_0^3+x_1^3+x_2^3)+24x_0x_1x_2$
find the number and the multiplicity of the intersections between $p\,$ and $q$.
For the Bézout's theorem the number of intersections counted with their multiplicity is the product of the degrees of the two polynomials, which is $9$. Now I want to use the main theorem of elimination theory to eliminate one of the coordinates, in order to work on a single polynomial of degree $9$ in two variables. But I can't apply it because $p(1,0,0)=p(0,1,0)=p(0,0,1)=0$ and I need that one of those points doesn't lie on both of the curves to eliminate the corresponding coordinate.
Any suggestions? Maybe an appropriate automorphism of $\mathbb{P^2(K)}$ to workaround.
I computed the elimination of $x_0$ and the polynomial I obtained has the right degree: $-64x_1^9+192x_1^6x_2^3+1536x_1^3x_2^6+64x_2^3 \, = \, -64(x_1^3-3x_1x_2^2-x_2^3)(x_1^6+3x_1^4x_2^2-2x_1^3x_2^3+9x_1^2x_2^4-3x_1x_2^5+x_2^6)$
but $x_1=0$,$\,\,x_2=0$ is a partial solution and it doesn't correspond to any solution of the system.
 A: First, suppose $x_0=0$.  Then $p(0,x_1,x_2)=x_1x_2^2$ and $q(0,x_1,x_2)=-8(x_1^3+x_2^3)$, which intersect at $x_1=x_2=0$.  Thus any intersection must occur on the affine chart $\{x_0\neq0\}$.  In fact, one can generalize by symmetry: the intersections lie in the chart $\{x_0x_1x_2\neq0\}$.  For this reason, I think your analysis of $p(1,0,0)$ etc. does not actually obstruct the main theorem.  But converting multiple equations to a single high-degree polynomial usually makes things worse, not better.
Instead, follow Avnish Singh's advice to divide both equations by $x_0x_1x_2$.  Then we have \begin{gather*}
0=\frac{x_1}{x_2}+\frac{x_2}{x_0}+\frac{x_0}{x_1} \\
0=8\left(3-\left(\frac{x_0}{x_1}\frac{x_0}{x_2}+\frac{x_1}{x_0}\frac{x_1}{x_2}+\frac{x_2}{x_0}\frac{x_2}{x_1}\right)\right)
\end{gather*}  Now change variables: let $(\alpha,\beta,\gamma)=\left(\frac{x_0}{x_1},\frac{x_1}{x_2},\frac{x_2}{x_0}\right)$.  Then \begin{gather*}
1=\alpha\beta\gamma \\
0=\alpha+\beta+\gamma \\
3=\frac{\alpha}{\gamma}+\frac{\beta}{\alpha}+\frac{\gamma}{\beta}
\end{gather*}
Now let $(l_0,l_1,l_2)=\left(\frac{\alpha}{\gamma},\frac{\beta}{\alpha},\frac{\gamma}{\beta}\right)$.  Then since $\gamma\neq0$, \begin{gather*}
1=l_0l_1l_2 \\
1=l_0^2l_1\gamma^3 \\
0=l_0+l_1l_0+1 \\
3=l_0+l_1+l_2
\end{gather*} which are easy enough to solve by hand.
A: Equating both sides, and then bringing 24*x$_0*$x$_1*$x$_2$ to the left side and then dividing both sides by x$_0$x$_1$x$_2$ gives us
$24- [\frac{x_0}{x_1} + \frac{x_1}{x_2} + \frac{x_2}{x_0}] = 8[\frac{x^2_0}{x_1*x_2} + \frac{x^2_1}{x_0*x_2} + \frac{x^2_2}{x_0*x_1}]$
Consider the first bracket terms as m and the second brackets term as n.
Now, 24 - m = 8n
Also, using AM $\ge$ GM, we can find out the minimum values of m and n to be 3.
Now take m = 3 + i where i $\ge$ 0
and n =3 + j where j $\ge$ 0.
Putting these values in the above equation, we get
$24 - 3 - i= 8(3+j) $
$\implies -i -8j= 3$
Now, where i and j are positive values , the above equation has no solutions.
Thus the solutions exist only when we can't equate AM $\ge$ GM
that is when $x_1$ or $x_2$ or $x_3$ = $0$
