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}
 A: Consider the Moebius strip with midline the circle of radius $R$ in the $(x,y)$-plane and having width $2a>0$:
$$M:\quad(\phi,s)\mapsto\left\{\eqalign{x&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi \cr
y&=\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi \cr
z&=s\sin{\textstyle{\phi\over2}}\ .\cr}\right.\tag{1}$$
The parameter domain is $-\pi<\phi<\pi$, $\ -a<s<a$. For "technical reasons" we have excluded $\phi=\pm\pi$. This guarantees $\cos{\phi\over2}>0$ over the whole parameter domain. 
It is claimed that $M$ is part of a certain cubic surface $S\subset{\mathbb R}^3$. This surface results from revoking the condition $-a<s<a$ in $(1)$, so that now the parameter $s$ runs from $-\infty$ to $\infty$. Consider a fixed value $\phi\in\ ]{-\pi},\pi[\ $. Then 
$$g_\phi:\quad s\mapsto\left(\bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\cos\phi, \ \bigl(R+s\cos{\textstyle{\phi\over2}}\bigr)\sin\phi , \ s\sin{\textstyle{\phi\over2}}\right)\tag{2}$$
describes a  straight line lying on $S$. We now replace the parameter $s$ in $(2)$ by the new parameter $u:=R+s\cos{\phi\over2}$. In this way $g_\phi$ appears in the form
$$g_\phi:\quad u\mapsto\bigl(u\cos\phi, u\sin\phi, \tan{\textstyle{\phi\over2}}(u-R)\bigr)\qquad(-\infty<u<\infty)\ .\tag{3}$$
When we let $\phi$ vary as well in $(3)$ we obtain another parametrization of our surface $S$. In order to get rid of the trigonometric functions we introduce the new parameter $t:=\tan{\phi\over2}$, which runs from $-\infty$ to $\infty$. In this way we obtain the following rational representation of $S$:
$$S:\quad(t,u)\mapsto\left\{\eqalign{x&=u\ {1-t^2\over 1+t^2} \cr
y&=u\ {2t\over1+t^2}\cr
z&=t\ (u-R)\cr}\right.\qquad\qquad\bigl((t,u)\in{\mathbb R}^2\bigr)\ .$$
One has $2(z+Rt)=2ut=y(1+t^2)$ and $$2tx=y(1-t^2)\ .\tag{4}$$ Adding these two equations leads to
$$t={y-z\over R+x}\ .$$
We insert this value of $t$ into $(4)$ and obtain after clearing denominators
$$2x(R+x)(y-z)=y\bigl((R+x)^2-(y-z)^2\bigr)\ .$$
This equation is valid for all points $(x,y,z)\in S$, and as well on the line $g_{\pm \pi}$ omitted from consideration. It expands to
$$-R^2 y + x^2 y + y^3 - 2 R x z - 2 x^2 z - 2 y^2 z + y z^2=0\ ,$$
as given in the quoted Wikipedia link (in your question you have forgotten one term).
A: 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.
