# What is the method to solve this system of equations?

$$\Huge{\text{Updated:}}$$

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.

• Groebner bases is the magic word. – Wuestenfux Sep 10 '19 at 12:44
• @Wuestenfux What is the Groebner..? Sorry English is not my native language. – Learner Sep 10 '19 at 12:46
• Basically its one of the triumphs of computational algebraic geometry. It allows you to solve systems of polynomial equations by elimination. – Wuestenfux Sep 10 '19 at 12:51
• @Wuestenfux Do I need to do this with the help of computer? – Learner Sep 10 '19 at 12:59

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.
• So, for $A = B = 1$ and for only some special nonzero values of $A$ and $B$ ,we can escape cubic equation. Do I understand correct? – Learner Sep 10 '19 at 13:29
• Teacher, I want to solve this system of equations for all non-zero $A$ and $B$ by avoiding the cubic equation. Which You've done before for $A=B=1$. – Learner Sep 10 '19 at 13:46
• I don't understand you, too :) I am afraid that we cannot avoid a polynomial equation (so not only cubic, but also perhaps quartic and higher) for general $A$ and $B$. If you satisfy some cubic in $A$ and $B$ like $A^2=B^3$ by setting, say, $A=B=1$, then it becomes possible, but not for general $A$ and $B$. – Dietrich Burde Sep 12 '19 at 15:19