Identity, Maths Club of IISER Kolkata

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. ...