Prove that determinant is zero iff $x= 2, 3$ 
Prove that
$$\begin{vmatrix}
1 & x & x^{2}\\ 
1 & 2 & 2^{2}\\ 
1 & 3 & 3^{2}
\end{vmatrix}=0\Leftrightarrow x= 2, 3$$

I see that if determinant is zero, two lines must equal, in this problem, there are between line 1 and line 2, 3, but how can I prove there is no other solutions ?? And what for $x^{3}, x^{4}, etc$ instead ?? Thks
 A: The Leibniz formula for determinants implies that the determinant in question is a quadratic polynomial in $x$. Since you have found two different roots, they are all the roots.
A: $$p(x) = \begin{vmatrix}
1 & x & x^{2}\\ 
1 & 2 & 2^{2}\\ 
1 & 3 & 3^{2}
\end{vmatrix}= (x-2)(x-3)$$
Why? See that $p(x)$ has degree at most $2$, and it has two roots $2,3$ (by inspection). So it must be of the form $\alpha(x-2)(x-3)$ for some $\alpha \in \mathbb{R}$. $\alpha = 1$, but you don't really need this information to conclude that $2$ and $3$ are the only roots of $p(x)$, since we know that any polynomial $f(x)$ has roots $x_1,...,x_n$, iff $\lambda f(x)$ has roots $x_1,...,x_n$ for some scalar $\lambda\in\mathbb{R}$.
It should be clear now!
As mentioned in the comments, you can look at Vandermonde determinants for a generalization.
A: Why not to expand the determinant and find all the solutions?
$$\begin{vmatrix}
1 & x & x^{2}\\ 
1 & 2 & 2^{2}\\ 
1 & 3 & 3^{2}
\end{vmatrix} = 
\begin{vmatrix}
1 & x & x^{2}\\ 
0 & 2-x & 2^{2}-x^2\\ 
0 & 1 & 5
\end{vmatrix} = 
\begin{vmatrix} 
2-x & 2^{2}-x^2\\ 
1 & 5
\end{vmatrix}$$
$$=10-5x-(4-x^2) = x^2-5x+6=(x-2)(x-3)$$
Which equals to zero iff $x=2,3$
