Derive Cartesian cubic Möbius strip from parametric The following link:
http://mathworld.wolfram.com/MoebiusStrip.html
shows the Möbius strip parametrized as
\begin{eqnarray}
x = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \cos t \\
y = [ R + s \cos \left ( \frac{1}{2} t \right ) ] \sin t \\
z = s \sin \left ( \frac12 t \right )
\end{eqnarray}
The symbols for $R$ and $s$ and angle $t$ are explained there.
Then they say that from this parametrization we can derive the cubic.
\begin{equation}
-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z + y z^2 = 0.
\end{equation}
Any ideas about how to do this? I have tried with no success.
Thanks.

This is what I have done so far:
Square the first two equations above and add to find
\begin{equation}
  x^2 + y^2 = \left ( R + s \left ( \cos \frac{t}{2} \right )  \right )^2
\end{equation}
Take the square root of this
\begin{equation}
  \sqrt{x^2 + y^2 }=  R + s \left ( \cos \frac{t}{2} \right )   \Longrightarrow
  \sqrt{x^2 + y^2 } - R =   s \left ( \cos \frac{t}{2} \right )  
\end{equation}
Square this and the third equation (for $z$) and add to find
\begin{equation}
  s^2 = \left ( \sqrt{x^2+y^2}- R  \right )^2 + z^2
\end{equation}
Now, let us divide the second by the first equation
That is
\begin{equation}
  \frac{y}{x} = \tan t = \frac{2 \tan (t/2)}{1 - \tan^2 (t/2)}
\end{equation}
multiply numerator and denominator by $\cos^2 (t/2)$
\begin{equation}
  \frac{y}{x} = \frac{2 \sin(t/2) (\sqrt{1-\sin^2(t/2)} }{\cos^2 (t/2) - \sin^2(t/2)}
  = \frac{2 \sin(t/2) \sqrt{1 - \sin^2(t/2)}}{1 -   2 \sin^2(t/2)}.
\end{equation}
Multiply numerator and denominator by $s^2$, then
\begin{equation}
  \frac{y}{x} = \frac{2 z  \sqrt{s^2 - z^2}}{s^2 - 2 z^2}
\end{equation}
That is
\begin{equation}
  \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ ( \sqrt{x^2 + y^2} - R)^2 - z^2}
\end{equation}
or
\begin{equation}
  \frac{y}{x} = \frac{2 z (\sqrt{x^2 + y^2} - R)}{ (x^2 + y^2) + R^2 - 2 R \sqrt{x^2+y^2} 
  - z^2}
\end{equation}
Or
\begin{equation}
  y (x^2 + y^2) + y R^2 - 2 R y  \sqrt{x^2+y^2}  - z^2 y = 
  2 z x \sqrt{x^2  + y^2} -2 R x z
\end{equation}

I'd just substitute the given coordinates into the equation and show that it's satisfied. The terms can be grouped according to the powers of $R$ and $s$ they contain, with the two exponents always summing to $3$, and the equation has to be satisfied for all four groups separately; that makes the calculation more manageable. For example, for $R^3s^0$ I get
$$-\sin t+\cos^2t\sin t+\sin^3t=0\;,$$
which is indeed satisfied.
