solve the following determinant
|1 a a^2|
|1 b b^2|
|1 c c^2|
add the opposite of row 1 to row 2 and 3 to get
$\displaystyle \begin{bmatrix}
1 & a & a^2 \\
0 & b-a & b^2-a^2 \\
0 & c-a & c^2 -a^2 \\
\end{bmatrix}$
Now we can expand along the first column to get
$\displaystyle 1\cdot ( (b-a)(c^2-a^a)-(c-a)(b^2-a^2)) = $
$\displaystyle (b-a)(c-a)(c+a)-(c-a)(b-a)(b+a)=[(b-a)(c-a)][(c+a)-(b+a)]=$
$\displaystyle (b-a)(c-a)(c-b)$