$\Huge{\text{Updated:}}$
In this question I got an answer for special non-zero values of $A$ and $B.$ I need to solve these polynomial equations for any arbitrary non-zero coefficients.
$\underline{\text{ I am looking for a solution that does not lead to the solution of a cubic equation.}}$
Here is the problem:
For all arbitrary non-zero coefficients $A$ and $B$, I am looking for a method to solve this system of equations, where $x\neq 0,y\neq 0,z\neq 0, u\neq 0, v\neq 0.$
$$\begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &(1)\\
3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B=0 &(2)\\
3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y=0 &(3)\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &(4)\end{cases} $$
$\Huge{\text{My old attempts:}}$
$3y^2+2vzyA+u^2yA+2xzuyA+y^2z^2A+3vz^2yB+3x^2y+3zu^2yB-3y^2-v^2A-2xuvA-2yzvA-3v^2zB-yu^2A-3vu^2B-3x^2y=0\Longrightarrow (yz-v)(2xuA+3vzB+yzA+vA+3u^2B)=0$
Let, $v=yz$ we get from $(1)$ and $(4)$
$yz^2A+zvA=0 \Longrightarrow zy+v=0 \Longrightarrow 2v=0 \Longrightarrow v=0$ which is contradiction. So, $yz\neq v$
and $2xuA+3vzB+yzA+vA+3u^2B=0$ must be.
$\Huge{\text{My new attempts:}}$
$z(2xuA+3vzB+yzA+vA+3u^2B)-(3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B)=0 \Longrightarrow 3y+zvA+u^2A+3x^2=0$
$v(2xuA+3vzB+yzA+vA+3u^2B)-(3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y)=0 \Longrightarrow 3y+zvA+u^2A+3x^2=0$
Finally, I can construct a new system of equations: (If I didn't make any mistake)
$$\begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &\\
3y+2vzA+u^2A+2xzuA+yz^2A+3vz^2B+3x^2+3zu^2B=0 &\\
3y^2+v^2A+2xuvA+2yzvA+3v^2zB+yu^2A+3vu^2B+3x^2y=0 &\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &\end{cases} \Longrightarrow \begin{cases}3z^2uB+3x+2uzA+xz^2A=0 &(1)\\
2xuA+3vzB+yzA+vA+3u^2B=0 &(2)\\
3y+zvA+u^2A+3x^2=0 &(3)\\
3xy^2+xv^2A+2yuvA+3v^2uB=0 &(4)\end{cases} $$
Here I am stuck. I've worked hard to make variables dependent on a single variable. Having a single variable requires a method. I can't find a method right now.
Best Answer
You can still eliminate. From $(1)$ we have $$ y = \frac{ - 3x^2 - A(u^2 + vz)}{3}, $$ and then from $(3)$ we have $$ x = \frac{uz( - 2A - 3Bz)}{Az^2 + 3}. $$ Actually, the case of $Az^2+3=0$ leads to $Bz^3-2=0$. Otherwise we have only two equations left in three variables $z,u,v$, namely $(2),(4)$ and can take the resultant. Then one can even specify something, e.g., $u=z$ and $v=-z$ to obtain a general solution. Here is one example: $$ u=z,\; v=-z\; \text{ with $z$ such that } Bz^3 + Az^2 +1=0. $$ Then $x=1$ and $y=-1$.
Edit: The general solution. After eliminating $x$ and $y$ as above we can eliminate $v$ by $$ v:= \frac{u^2(A^4z^5 + 30A^3z^3 + 45A^2Bz^4 + 45A^2z + 27ABz^5 - 81B}{- A^4z^6 - 3A^3z^4 + 9A^2Bz^5 + 9A^2z^2 + 54ABz^3 + 27A + 81Bz}, $$ provided the denominator is nonzero. In this case we obtain a single equation, which gives all solutions, namely $$ (AZ^2 + Bz^3 + 1)(2A^6z^6 + 90A^5z^4 + 162A^4Bz^5 + 270A^4z^2 + 108A^3B^2z^6 + 540A^3Bz^3 + 54A^3 + 1215A^2B^2z^4 - 486A^2Bz + 1458AB^3z^5 + 729B^4z^6 + 729B^2)=0. $$ Note that this is independent of the variable $u$. In case that the denominator is zero, we have a special case "avoiding the cubic", but introducing a dependency of $A$ and $B$, which was not allowed.