Solving system of equations: $x^3-3y^2x=-1$ and $3yx^2 -y^3=1$

Question

$$x^3-3y^2x=-1$$ $$3yx^2 -y^3=1$$

This was the real part and imaginary part on a previous question I asked, instead of the system it was easier to just use polar coordinates to solve, but if this was just a system unrelated to the problem, how would one solve it? I seem to have trouble when the terms are like this. All I can see is that x and y are nonzero so we can divide by them.

How would you approach this system?

I have $$x(x^2-3y^2)=-1$$ $$y(3x^2 -y^2)=1$$

Answer 1

Eliminating $y$, we have $$y^2=\frac{x^3+1}{3x}\ ,\quad y(3x^2-y^2)=1$$ and so $$\frac{x^3+1}{3x}\Bigl(3x^2-\frac{x^3+1}{3x}\Bigr)^2=1\ .$$ Multiplying both sides by $27x^3$, the equation becomes a cubic in $x^3$. The cubic happens to have one rational root, so all roots can easily be found.

Answer 2

HINT:

$$x^3+3x^2y-3xy^2-y^3=0$$

$$(x-y)(x^2+xy+y^2)+3xy(x-y)=0\iff(x-y)(x^2+xy+y^2+3xy)=0$$

Check both cases

Answer 3

If you want to use complex numbers, here's an easier ad hoc way (I tried to describe how I found the solution step by step):

Note that, the terms involved arise in the expansion of $(x-y)^3$, but the signs are messed up. In particular, we need signs of the terms in the same equation to be same to use the formula.

$i^2=-1$ can be used to flip the signs. So, multiply first equation by $i^3$ and second by $i^2$. On adding the equations, LHS becomes $(ix+y)^3$. Solve using polar representation.

Answer 4

• When $x = y$ the system reduces to $ x = y= t $ and $ 2t^3 = 1 $ hence the solution is $x = y= \dfrac{1}{\sqrt[3]2}$
• Suppose $x \neq y$ then the system reduces to (after adding both equations) $$ x^3 - y^3 = 3y^2x - 3yx^2 $$

$$ \require{cancel} \cancel {(x - y)} (x^2 + xy + y^2) = - 3xy \cancel {(x - y)} $$

$$ x^2 + 4xy + y^2 = 0 $$

$$ [x + (2 - \sqrt 3) y] [x + (2 + \sqrt 3) y ] = 0 $$

Hence the set of solutions is;

$$ (x, y ) \in \{ (\frac{1}{\sqrt[3]2}, \frac{1}{\sqrt[3]2}) \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x = (\sqrt 3 - 2) y \} \bigcup \{ (x, y) \ | \ x \neq y \; \text{and} \; x = - (\sqrt 3 + 2) y \} $$

Answer 5

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Lets $\ds{3yx^{2} \equiv \cosh^{2}\pars{t}}$ and $\ds{y^{3} \equiv \sinh^{2}\pars{t}}$ such that

\begin{align} y&=\sinh^{2/3}\pars{t}\,, \qquad x={\cosh\pars{t} \over \root{3}y^{1/2}} ={\cosh\pars{t} \over \root{3}\sinh^{1/3}\pars{t}} \end{align}

\begin{align} -1&={\cosh^{3}\pars{t} \over 3\root{3}\sinh\pars{t}} - 3\sinh^{4/3}\pars{t}\, {\cosh\pars{t} \over \root{3}\sinh^{1/3}\pars{t}} \\[5mm]&={\root{3} \over 9}\,{\cosh^{3}\pars{t} \over \sinh\pars{t}} - \root{3}\sinh\pars{t}\cosh\pars{t} \end{align}

\begin{align} 9\sinh\pars{t}&=\root{3}\bracks{9\sinh^{2}\pars{t} - \cosh^{2}\pars{t}}\cosh\pars{t} \\[5mm]&=\root{3}\bracks{8\sinh^{2}\pars{t} - 1}\root{\sinh^{2}\pars{t} + 1} \end{align}

With $\ds{z \equiv \sinh^{2}\pars{t}}$ we'll have: \begin{align} 27z&=\pars{8z - 1}^{2}(z + 1)\ \imp\ 64z^{3} + 48z^{2} - 42z + 1=0\ \imp\ \left\{\begin{array}{rcl} z_{0} & = & \phantom{-\,}\half \\[2mm] z_{1} & = & -\,{3\root{3} + 5 \over 8} \\[2mm] z_{2} & = & \phantom{-\,}{3\root{3} - 5 \over 8} \end{array}\right. \\[5mm]\mbox{and}&\qquad x={\root{3} \over 3}\,{\root{1 + z} \over z^{1/6}}\,,\qquad y=z^{1/3} \end{align}