Solving systems of equations in 3 variables

Question

I am trying to solve (in $\mathbb{C}$) the following systems of equations in 3 variables:

$$x+y+z=0$$ $$xy+yz+xz=c-c^2$$ $$xyz=3c^2$$ where, $c$ is some fixed non-zero constant.

I tried substituting in $x=-y-z$ into the third equation, and then solved for $z$, but the problem is that this is not a nice answer and so substituting this into the second equation leads to a disaster.

I am assuming there is a neat trick as this looks like a special case of systems of equations in 3 variables!?

Is it also correct to assume that there is a unique solution for $x,y,z$ since there are 3 equations and 3 unknowns?

Answer 1

Write your system in the form $$xy-(x+y)^2=c-c^2$$ and $$-xy(x+y)=3c^2$$ Substituting $$x+y=a$$ $$xy=b$$ so you will get $$b-a^2=c-c^2$$ and $$-ab=3c^2$$ Can you proceed? Eliminating $b$ you will get for $a$: $$0=a^3+a(c-c^2)+3c^2$$

Answer 2

Let $z=-x-y$. Substitute this into the second and third equation. Multiplying the second by $y$ and subtracting the third we obtain $$ 2xy^2 + c(cy + 3c - y)=0. $$ For $y=0$ we have $x=y=c=0$. Otherwise we can substitute $x=-c(cy+3c-y)/(2y^2)$. This yields only one polynomial equation in $y$ and $c$, namely, $$ c^3( - y^2 - 6y - 9) + 2c^2y(y + 3) + cy^2(2y^2 - 6y - 1) - 2y^4=0. $$ I doesn't get easier than that. But it can be solved over the complex numbers. One particular solution is $c=1$ and $2y^3+3=0$, $z=3/y^2$, $x=-3/(2y^2)$. However, it is not correct that a non-linear system in $n$ variables with $n$ equations has a unique solution.

Answer 3

There is a classical 'trick' one can use for these types of systems of equations, namely Vieta's formulas. Note that $$ (\lambda - x)(\lambda - y)(\lambda - z) = \lambda^3 - (x + y + z) \lambda^2 + (xy +yz + xz)\lambda - xyz. $$ Hence, $x, y, z$ are the three roots of the polynomial equation $$ \lambda^3 + (c - c^2) \lambda - 3c^2 = 0. $$ Denoting the roots of this equation by $\lambda_1,\lambda_2, \lambda_3$, this shows that the only solutions of the original system of equations are permutations of $(\lambda_1, \lambda_2, \lambda_3)$. Moreover, as there is an explicit (albeit rather complicated) expression for the roots of a cubic polynomial, one can find the solutions of the above equation.