Co-planar points on the twisted cubic

Here's a fun problem from a course on Curves and Surfaces.

Consider the following curve, called the twisted cubic. \[ \gamma\colon \mathbb{R} \to \mathbb{R}^3,\qquad \gamma(t) = (t, t^2, t^3). \] Show that no four distinct points on this curve can lie on the same plane.

Method 1 - The scalar triple product

The scalar triple product is a good tool to have at one's disposal (thanks to Satbhav Voleti for sharing this solution); given three vectors $\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}$, their scalar triple product $\boldsymbol{u}\cdot(\boldsymbol{v}\times \boldsymbol{w})$ represents the volume of the parallelepiped whose legs are these vectors. Now if these three vectors all lie in the same plane, the corresponding parallelepiped is completely flat with zero volume, thus the scalar triple product is also zero. Coming back to our problem, suppose that the four distinct points on the twisted cubic $P_i = (t_i, t_i^2, t_i^3)$, $i = 0, 1, 2, 3$, all lie on the same plane. This means that the difference vectors $P_1 - P_0$, $P_2 - P_0$, $P_3 - P_0$ have a zero scalar triple product. Luckily, this quantity is easy to compute; it is simply the determinant of the matrix whose columns are the three vectors. This means that we want \[ \det\begin{bmatrix}t_1 - t_0 & t_2 - t_0 & t_3 - t_0 \\ t_1^2 - t_0^2 & t_2^2 - t_0^2 & t_3^2 - t_0^2 \\ t_1^3 - t_0^3 & t_2^3 - t_0^3 & t_3^3 - t_0^3 \end{bmatrix} = 0. \] This particular determinant is somewhat well known (as we will see in the next method), and this simplifies to \[ (t_1 - t_0)(t_2 - t_0)(t_3 - t_0)(t_2 - t_1)(t_3 - t_1)(t_3 - t_2) = 0.\] With distinct $t_0, t_1, t_2, t_3$, this simply cannot happen.

Method 2 - Simultaneous linear equations

Suppose that four distinct points $(t_i, t_i^2, t_i^3)$, $i = 0, 1, 2, 3$ lie on the plane defined by $k + ax + by + cz = 0$. We can plug in the values of $x, y, z$ for each of the four points, and obtain a system of linear equations summarized as \[\begin{bmatrix}1 & t_0 & t_0^2 & t_0^3 \\ 1 & t_1 & t_1^2 & t_1^3 \\ 1 & t_2 & t_2^2 & t_2^3 \\ 1 & t_3 & t_3^2 & t_3^3\end{bmatrix}\begin{bmatrix}k\\a\\b\\c\end{bmatrix} = \boldsymbol{0}.\] The matrix on the left is called a Vandermonde matrix; in fact, its determinant is precisely what we computed earlier, and is non-zero. Thus, it is invertible, and multiplying by this inverse gives us the only solution, $k = a = b = c = 0$. In other words, there is no common 2D plane.

Method 3 - The cubic

This is perhaps the quickest, and definitely my favourite solution here. Like before, suppose that four distinct points $(t_i, t_i^2, t_i^3)$, $i = 0, 1, 2, 3$ lie on a (proper) plane $k + ax + by + cz = 0$. Plugging in our points, this would imply that the non-zero cubic $k + at + bt^2 + ct^3$ has four distinct roots, which is absurd.

It is interesting to note that this fact (a non-zero polynomial of degree $n$ can have at most $n$ distinct roots) can be demonstrated/proved using Vandermonde matrices, just as we have done.

Here we have seen solutions involving geometry, linear algebra, and good old 'regular' algebra. Can you think of any other ways of solving this problem?