Is there a solution for x for $x^2+(y-\sqrt[3]{x^2})^2=1$?

Question

I tried to solve this equation (heart shaped function) for x, but cannot get a nice solution. Why?

$x^2+(y-\sqrt[3]{x^2})^2=1$

The formula is taken from this image:

I just wanted to rewrite the formula in order to be able to solve by $x$.

Answer 1

What exactly do you mean by solve? You can put it in the form $$ y = x^\frac{2}{3} \pm \sqrt{1 - x^2} $$ or $$ y = |u|^3 \pm \sqrt{1 - |u|} $$ where $u = x^\frac{3}{2}$, but I don't think there is a nice solution in the form x = f(y), since this involves finding the inverse of a third degree polynomial.

Answer 2

$$x^2+\left(y-\sqrt[3]{x^2}\right)^2=1\Longleftrightarrow$$ $$\left(y-\sqrt[3]{x^2}\right)^2=1-x^2\Longleftrightarrow$$ $$y-\sqrt[3]{x^2}=\pm\sqrt{1-x^2}\Longleftrightarrow$$ $$y=\pm\sqrt{1-x^2}+\sqrt[3]{x^2}\Longleftrightarrow$$ $$y=\sqrt[3]{x^2}\pm\sqrt{1-x^2}$$

If you want to solve $x$ (get $x$ as function of $y$) it becomes very messy!

Answer 3

To get an idea of what you're asking for, here is the first solution:

$$x=-\sqrt{-y^2+\frac{14 \sqrt[3]{2} y^3}{\sqrt[3]{-216 y^5-783 y^4-450 y^3+144 y^2+\sqrt{4 \left(-42 y^3-42 y^2+6 y+5\right)^3+\left(-216 y^5-783 y^4-450 y^3+144 y^2+126 y-11\right)^2}+126 y-11}}+\frac{14 \sqrt[3]{2} y^2}{\sqrt[3]{-216 y^5-783 y^4-450 y^3+144 y^2+\sqrt{4 \left(-42 y^3-42 y^2+6 y+5\right)^3+\left(-216 y^5-783 y^4-450 y^3+144 y^2+126 y-11\right)^2}+126 y-11}}-\frac{2 \sqrt[3]{2} y}{\sqrt[3]{-216 y^5-783 y^4-450 y^3+144 y^2+\sqrt{4 \left(-42 y^3-42 y^2+6 y+5\right)^3+\left(-216 y^5-783 y^4-450 y^3+144 y^2+126 y-11\right)^2}+126 y-11}}+\frac{\sqrt[3]{-216 y^5-783 y^4-450 y^3+144 y^2+\sqrt{4 \left(-42 y^3-42 y^2+6 y+5\right)^3+\left(-216 y^5-783 y^4-450 y^3+144 y^2+126 y-11\right)^2}+126 y-11}}{3 \sqrt[3]{2}}-\frac{5 \sqrt[3]{2}}{3 \sqrt[3]{-216 y^5-783 y^4-450 y^3+144 y^2+\sqrt{4 \left(-42 y^3-42 y^2+6 y+5\right)^3+\left(-216 y^5-783 y^4-450 y^3+144 y^2+126 y-11\right)^2}+126 y-11}}-2 y+\frac{2}{3}}$$