Let $A=\pmatrix{1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\\ 1 & \gamma & \gamma^2}$ Furthermore we have the function $p$:
$p:\mathbb{R}\to\mathbb{R}$,
$x \mapsto \det\pmatrix{1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\\ 1 & x & x^2}$, $\alpha\neq \beta$.
My objective is to prove that $p$ is a polynomial that is different from the zero polynomial and to determine its roots.
I get that: $\det(A) = \; (\alpha - \beta) \; (\alpha - \gamma) \; (\gamma - \beta)$.
I can conclude that $A$ is invertible if and only if $\alpha$,$\beta$ and $\gamma$ are all distinct.
I also observe that if $A$ is row-equivalent to $I_3$, then it's invertible.
Before row reducing $A$, don't I have to assume that $\det(A) \neq 0$ (and therefore basically assuming that $\alpha ,\beta$ and $\gamma$ are pairwise different from each other, which is what I want to prove)?
How do I proceed?
You don't have to assume anything (in principle) to do row reduction. You have $$ \begin{bmatrix}1 & \alpha & \alpha^2\\ 1 & \beta & \beta^2\\ 1 & \gamma & \gamma^2\end{bmatrix} \to \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & \beta -\alpha& \beta^2-\alpha^2\\ 0 & \gamma-\alpha & \gamma^2-\alpha^2\end{bmatrix}. $$ If $\beta-\alpha=0$, the second row is zero. If $\gamma-\alpha=0$, the third row is zero. In either case, $A$ is not invertible. When $(\beta-\alpha)(\gamma-\alpha)\ne0$, you can continue the row reduction $$ \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & \beta -\alpha& \beta^2-\alpha^2\\ 0 & \gamma-\alpha & \gamma^2-\alpha^2\end{bmatrix}\to \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & 1& \beta+\alpha\\ 0 & 1 & \gamma+\alpha\end{bmatrix} \to \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & 1& \beta+\alpha\\ 0 & 0 & \gamma-\beta\end{bmatrix} $$ Now if $\gamma-\beta=0$ you are done and $A$ is not invertible; if $\gamma-\beta\ne0$, you can continue $$ \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & 1& \beta+\alpha\\ 0 & 1 & \gamma-\beta\end{bmatrix} \to \begin{bmatrix}1 & \alpha & \alpha^2\\ 0 & 1& \beta+\alpha\\ 0 & 1 & 1\end{bmatrix} \to I_3. $$