Can we represent a symmetric curve by a parameter with symmetry? 
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?
 A: 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.
