<s>[INST] 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
 [/INST] 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 $x=2,3$
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