Let $C(F(x,y))$ denote the zero locus of $F(x,y)$
Let $A=${$(x,y) : y\ge\frac{2\sqrt{3}}{3}$}
Let $B=${$(x,y) : 0\le y\le\frac{2\sqrt{3}}{3}$}
Let $C=${$(x,y) : -\frac{\sqrt{3}}{3}\le y\le0$}
Let $D=${$(x,y) : y\le-\frac{\sqrt{3}}{3}$}
It can easily be checked that:
- $$C(\sqrt[3]{\frac{x^{2}+\sqrt{x^{4}-\frac{4}{27}}}{2}}+\sqrt[3]{\frac{x^{2}-\sqrt{x^{4}-\frac{4}{27}}}{2}}-y)=C(y^3-x^2-y) \cap A$$
- $$C(-\frac{\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)+\sin\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap C$$
- $$C(-\frac{\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-\sin\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap D$$
By the fundamental theorem of algebra, $$y^3-y-x^2=(y-f(x))(y-g(x))(y-h(x))$$ for some $f,g,h$
Thus, I was expecting the zero locus of $y^3-x^2-y$ to be the union of the respective zero loci of 3 expressions of the form $y-f(x)$
What seems to be my misunderstanding?
Additionally, does there exist an expression $y-G(x)$ such that
$$C(G(x)-y)=C(y^3-x^2-y) \cap B$$
?
If so, how can it be derived?
It turns out that the zero locus of $y^3-x^2-y$ is indeed the union of the respective zero loci of 3 expressions of the form $y-f(x)$
It turns out that there is an expression $y-H(x)$ such that $$C(H(x)-y)=C(y^3-x^2-y) \cap (A\cup B)$$
Indeed,$$C(\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)-y)=C(y^3-x^2-y) \cap (A\cup B)$$
However, even though $\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)$ is real for $x\in (-\infty,-\sqrt[4]{\frac{4}{27}}]\cup [\sqrt[4]{\frac{4}{27}},\infty)$, we will have an issue with regard to desmos, as desmos cannot handle any expression involving the square root of a negative number.
This can be remedied, though. Note that $$\frac{2\sqrt{3}}{3}\cos\left(\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{12-81x^{4}}}{9x^{2}}\right)\right)=\frac{2\sqrt{3}}{3}\cosh\left(\frac{1}{3}\tanh^{-1}\left(\frac{\sqrt{81x^{4}-12}}{9x^{2}}\right)\right)$$
Thus, in order to get desmos to play nice with us, we can just use both of these, while remembering that in reality either one of these would be all that's needed for the entirety of $C(y^3-x^2-y) \cap (A\cup B)$