Proving Linearly Independence from System of Equation

Question

I am trying to understand the proof of Linearly Independence of the basis set $\{1, x, x^2, x^3\}$. It is written that -

Substituting $3$ other values of $x$ into the above equation yields a system of $3$ linear equations in the remaining $3$ unknowns $c_1, c_2$ and $c_3$.

The equation we are considering right now is - $$c_0 \cdot (1) +c_1x+c_2x^2+c_3x^3=0 \cdots (1)$$

First we plug in $x= 0$ to get the value of $c_0$ in equation $(1)$, we get -

$$ c_0 \cdot (1) +c_1\cdot 0 +c_2\cdot 0 +c_3\cdot 0=0\implies c_0 =0$$ For arbitrary 3 non-zero values $ x_1, x_2, x_3 \in \mathbb{R}$ where $ x_1, x_2, x_3$, are not solutions of equation $(1)$ and for $c_0=0$, we will get $3$ equations from equation $(1)$-

$$c_1x_1+c_2x_1^2+c_3x_1^3=0$$ $$c_1x_2+c_2x_2^2+c_3x_2^3=0$$ $$c_1x_3+c_2x_3^2+c_3x_3^3=0$$

Now, how can it be shown that the only solution of the above system of equations is the trivial solution $c_1 = c_2 = c_3 = 0$ and therefore the set $\{1, x, x_2, x_3\}$ is linearly independent?

Thanks.

The source of the problem and background is given below -

Answer 1

The matrix of that system is$$\begin{bmatrix}x_1&x_1^{\,2}&x_1^{\,3}\\x_2&x_2^{\,2}&x_2^{\,3}\\x_3&x_3^{\,2}&x_3^{\,3}\end{bmatrix}.$$It is a Vandermonde matrix and its determinant is $(x_1-x_2)(x_1-x_3)(x_2-x_3)$. So, if the numbers $x_1$, $x_2$, and $x_3$ are distinct, the only solution of the system is $(0,0,0)$, since then the determinant is not $0$.

Answer 2

An alternative approach: one might notice that $c_0+c_1x+c_2x^2+c_3x^3$ is a cubic polynomial and has at most three roots because of the fundamental theorem of algebra. Thus the only way this polynomial is $=0 ~ \forall x$ is if $c_0=c_1=c_2=c_3=0$.