Let $k_1$, $k_2$ be two real numbers such that the following system of $4$ variables $x_1$, $x_2$, $x_3$ and $x_4$ is satisfied:
$$k_1(x_1^3-x_2^3)+k_2(x_1^3-x_3^3)=x_1^3-x_4^3$$ $$k_1(x_1^2-x_2^2)+k_2(x_1^2-x_3^2)=x_1^2-x_4^2$$ $$k_1(x_1-x_2)+k_2(x_1-x_3)=x_1-x_4$$
Prove that $x_1$, $x_2$, $x_3$ and $x_4$ can't be all different, that is $x_i$ must equal $x_j$ for some $i \neq j$, where $ 1\leq i,j \leq4 $.
I have made that problem up -kinda off-, and I know that the result is true, but I can't prove it.
I'm more interested in learning a systemic way to approach those kind of problems (problems of solving non-linear equation), more than proving this result in particular -although it would be nice to see a clean proof-.
By the way, I have stumbled on this problem while I'm trying to prove that a polynomial equation of degree 3, can't have more than 3 solutions. I have constructed the system by writing the third raw of a matrix (with det=0) as a linear combination of the first tow rows.
So how I can approach the problem?
Let $X = \begin{pmatrix} x_1^3 - x_2^2 & x_1^3 -x_3^3 & x_1^3 - x_4^3 \\ x_1^2 - x_2^2 & x_1^2 - x_3^2 & x_1^2 - x_4^2 \\ x_1 - x_2 & x_1 - x_3 & x_1 - x_4 \end{pmatrix}.$
The existence of $k_1$ and $k_2$ means the last column is a linear combination of the first two.
Show that $\det(X) = (x_1 - x_2)(x_1 - x_3)(x_1 - x_4)(x_2 - x_3)(x_2 - x_4)(x_3-x_4)$.
So the determinant is non-zero if all the x's are pairwise distinct.
So if all x's are pairwise distinct, the matrix $X$ is non-singular, in particular, the last column cannot be a non-trivial linear combination of the first two. So if x's are pairwise distinct, we must have $k_1 = k_2 = 0$, but then, the last column must be zero and we must have $x_1 = x_4$ a contradiction.