<s>[INST] Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix} 
1 && 1 && 1 \\ 
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \end{pmatrix}$ - 
det$\begin{pmatrix}
a && c \\
a^2 && c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a && b \\
a^2 && b^2 \end{pmatrix}$
which gives 
$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
which simplifies to
$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$ 
and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.
 [/INST] It might be useful to look for helpful column/row operations before expanding.
Here, subtraction of the first column from the second and the third one simplify the determinant considerably:
$$\det\begin{pmatrix} 
1 && 1 && 1 \\ 
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = 
\det \begin{pmatrix} 
1 && 0 && 0 \\ 
a && b-a && c-a \\
a^2 && b^2-a^2 && c^2-a^2 \end{pmatrix} = \ldots $$
$$ \ldots =(b-a)(c-a)\det\begin{pmatrix}
1 && 1 \\
b+a && c+a \end{pmatrix} = (b-a)(c-a)(c-b)$$
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