Representing rational quadratic Bezier with non-rational cubic Is it possible to exactly represent a rational quadratic Bezier curve with a non-rational cubic Bezier curve? I know that it is always possible to exactly represent an n-degree Bezier with an (n+1)-degree Bezier by carefully choosing the control points, so I'm wondering if a similar technique can be applied to the rational-to-non-rational question.
My feeling is that it should be possible, since we are given an extra control point to work with. But, I also understand that the rational representation effectively lies in a higher dimensional space, which may not be the same thing.

As you probably know, a quarter-circular arc can be represented exactly with a quadratic rational Bézier curve. A cubic Bézier curve can only approximate that arc (quite well, but not exactly).
This shows that there are quadratic rational Bézier curves that cannot be re-parameterized as cubic Bézier curves.
Details:
Here is a parametrization of the quarter-circular arc around the origin
in the first quadrant:
$$\begin{align}
    x(t) &= \frac{1-t^2}{1+t^2}
&   y(t) &= \frac{2t}{1+t^2}
&   t&\in[0,1]
\end{align}$$
The same parametrization can be achieved with a rational quadratic Bézier curve
using the following control points and (non-normalized) weights:
$$\begin{align}
    P_0 &= (1,0)
&   P_1 &= (1,1)
&   P_2 &= (0,1)
\\  w_0 &= 1
&   w_1 &= 1
&   w_2 &= 2
\end{align}$$
If instead you want a circular arc around the origin given by real
polynomials $x(t)$ and $y(t)$, those would need to fulfill
$$\forall t\in[0,1]\colon\ x^2(t) + y^2(t) = 1\tag{*}$$
So clearly, those polynomials cannot be both zero.
Let $d$ be the maximum of the degrees of both polynomials.
Then the left-hand side of $(*)$ is a polynomial in $t$ of degree $2d$
with strictly positive leading coefficient.
Making the left-hand side a nonzero constant would therefore require $d=0$,
but that would yield only a point, not a curve segment.
By employing affine transformations, this result can be extended to
any elliptical arc anywhere in the plane: Polynomial Bézier curves cannot
parameterize it exactly, no matter how large the polynomial degrees involved.
