Question : Can we represent the following curve $C$ by one parameter $t$ as $x=f(t),y=g(t),z=h(t)$ with symmetry?
The curve $C$ in the $xyz$ space is defined as $$\begin{cases} x^2+y^2+z^2=1 \\ x^3+y^3+z^3=0 \\ \end{cases}$$ where $x,y,z\in\mathbb R.$
Motivation : I've already got the following example without symmetry :
For $0\le t\lt 2\pi$, $$f(t)=\frac{\cos t}{i(t)}, g(t)=\frac{\sin t}{i(t)}, h(t)=\frac{\sqrt[3]{\cos^3t+\sin^3t}}{i(t)}\ \text{where $i(t)=\sqrt{1+(\cos^3 t+\sin^3 t)^{2/3}}$}.$$
Since the given equations are symmetrical, I expect that there would exist an example with symmetry. Then, I reached the above question, but I'm facing difficutly. Can anyone help?
I've just been able to get an example.
First, note that $$\sin t+\sin\left(t+\frac{2\pi}{3}\right)+\sin\left(t+\frac{4\pi}{3}\right)=0.$$ For any $r\in\mathbb R, 0\le t\lt 2\pi$, letting $$x=r\sqrt[3]{\sin t}, y=r\sqrt[3]{\sin\left(t+\frac{2\pi}{3}\right)},z=r\sqrt[3]{\sin\left(t+\frac{4\pi}{3}\right)},$$ we know that these satisfy $x^3+y^3+z^3=0.$ Also we know that every point on $x^3+y^3+z^3=0$ can be represented by $r$ and $t$ in this way. Substituting these for $x^2+y^2+z^2=1$, we get $$x=\frac{\sqrt[3]{\sin t}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}},$$ $$y=\frac{\sqrt[3]{\sin\left(t+\frac{2\pi}{3}\right)}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}},$$ $$z=\frac{\sqrt[3]{\sin\left(t+\frac{4\pi}{3}\right)}}{\sqrt{(\sin t)^{2/3}+\left(\sin\left(t+\frac{2\pi}{3}\right)\right)^{2/3}+\left(\sin\left(t+\frac{4\pi}{3}\right)\right)^{2/3}}}$$ as desired.