Proving P∩Q=R∩S for Curves $P,Q,R,S$

[anemone]

Curves $P,\,Q,\,R,\,S$ are defined in the plane as follows:

$P=\{(x,\,y):x^3-3xy^2+3y=1\}$

$Q=\{(x,\,y):3x^2y-3x-y^3=0\}$

$R=\left\{(x,\,y):x^2-y^2=\dfrac{x}{x^2+y^2} \right\}$

$S=\left\{(x,\,y):2xy+\dfrac{y}{x^2+y^2}=3 \right\}$

Prove that $P\cap Q=R\cap S$.

 

[Albert1]

Curves $P,\,Q,\,R,\,S$ are defined in the plane as follows:

$P=\{(x,\,y):x^3-3xy^2+3y=1\}$

$Q=\{(x,\,y):3x^2y-3x-y^3=0\}$

$R=\left\{(x,\,y):x^2-y^2=\dfrac{x}{x^2+y^2} \right\}$

$S=\left\{(x,\,y):2xy+\dfrac{y}{x^2+y^2}=3 \right\}$

Prove that $P\cap Q=R\cap S$.

Spoiler

if we want to prove that: $P\cap Q=R\cap S$
then the solutions of $P\,\,and \,\, Q\,\, must\,\,=$ the solutions of $R \,\, and \,\,S$
to find their solutions is tedious,I will use the following approach:
from $R:$ $x^4-y^4=x---(1)$
from $P:$ $x^4-3x^2y^2+3xy=x---(2)$
from (1) and (2):$x^4-3x^2y^2+3xy=x^4-y^4$
or $3x^2y-3x-y^3=0 (y\neq 0)$
$\therefore$ from $ P,R$ we get equation $Q$
using the smilar method
from $ R,S$ we get equation $Q$
from $ P,Q$ we get equation $R$
--------
and the proof can be done this way

 

[anemone]

Bravo, Albert! I have never thought of approaching it like that! Good job!(Clapping)

Having said that, here is a solution of other that I also want to share with MHB:

Spoiler

Let $z=x+yi$.

The equations defining $P$ and $Q$ are the real and imaginary parts of the equation $z^3-3iz=1$ and similarly, the equations defining $R$ and $S$ are the real and imaginary parts of the equation $z^2=z^{-1}+3i$. Hence, for all real $x$ and $y$, we have

$(x,\,y)\in P\cap Q$ and $(x,\,y)\in R\cap S$

Thus $P\cap Q=R\cap S$.