Is $\mathbf{V}(\mathbf{X})=\mathbf{X^2}\times\mathbf{X}$ injective? I asked this question here a fortnight ago and answered it myself after pondering on it. Answers are still very much welcome and appreciated there, my answer is just the best I could come up with.
I took the vector $\mathbf{X}=(x_1,x_2,x_3)$, and the related $\mathbf{X^2}=(x_1^2,x_2^2,x_3^2)$, where the $x_t\in\Bbb R/\{0\}$ are not equal. 
I could then obtain $\mathbf{V}=\mathbf{X^2}\times\mathbf{X}$ for the simple solution that $c$ solves $$\mathbf{V}\cdot(\mathbf{1}c-\mathbf{Y})=0$$
For clarity $\mathbf{1}$, $c$ and $\mathbf{Y}$ are defined in the question linked at the top of this question - that's the context of this related question
I think that $\mathbf{V}$ is injective, that is, no two $\mathbf{X}$ map to the same $\mathbf{V}$, but I have no idea how to prove this, the algebra seems to go round in circles. 
Help would be appreciated - thanks in advance.
 A: Let ${\bf i, j, k}$ be the standard unit vectors of the space $\Bbb R^3$. Namely, ${\bf i}=(1,0,0)$, ${\bf j}=(0,1,0)$, and ${\bf k}=(0,0,1)$. Then $\bf{X^2}\times\bf{X}$ formally equals to a determinant 
$\left|\begin{matrix} \bf{i} & \bf{j} & \bf{k}\\
x_1 & x_2 & x_3\\
x_1^2 & x_2^2 & x_3^2
\end{matrix}\right|$. In particular, if $x_3=1$ then ${\bf V}=(x_2-x_2^2, x_1^2-x_1, x_1x_2(x_2-x_1))$. The first two coordinates of $\bf{V}$ are preserved by substitutions $x_1\to 1-x_1$ and $x_2\to 1-x_2$. So to show non-injectivity of $\bf{V}$ it suffices to find $x_1\ne x_2$ distinct also from $1$ and $1/2$ such that $x_1x_2(x_2-x_1)=(1-x_1)(1-x_2)(1-x_2-(1-x_1))$ that is $-x_1x_2=(1-x_1)(1-x_2)$. It suffices to have $x_2=\tfrac{1-x_1}{1-2x_1}$. For instance, we can pick $x_1=\tfrac 32$ and $x_2=\tfrac 14$. Then we have that ${\bf V}\left(\tfrac 32, \tfrac 14,1\right)= {\bf V}\left(-\tfrac 12, \tfrac 34,1\right)$.
PS. Even in $\Bbb R^2$ a quadratic is determined by five its points $(x_i, y_i)$, provided no four of them are collinear (see, for instance, [KK, 2.4.11]). In this case the equation of the quadratic is 
$$\left|\begin{matrix}
x^2 & xy & y^2 & x & y & 1\\
x_1^2 & x_1y_1 & y_1^2 & x_1 & y_1 & 1\\
x_2^2 & x_2y_2 & y_2^2 & x_2 & y_2 & 1\\
x_3^2 & x_3y_3 & y_3^2 & x_3 & y_3 & 1\\
x_4^2 & x_4y_4 & y_4^2 & x_4 & y_4 & 1\\
x_5^2 & x_5y_5 & y_5^2 & x_5 & y_5 & 1
\end{matrix}\right|=0.$$
The quadratic is degenerated iff three of the given points are collinear.
References 
[KK] Granino Korn, Theresa Korn Mathematical Handbook for scientists and engineers, 2nd edition, McGraw Hill, 1968 (Russian translation, Moskow, “Nauka”, 1973).
