How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped? The problem is as follows:

The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.
Prove this determinant
$\left|\begin{matrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$

What I've attempted to do was to use this well known identity on matrices:
$\left|\begin{matrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$
Then working in the right side of this equation I'm getting:
(Here I'm using the notation of capital $V$ for Vandermonde's determinant)
$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
Factoring the common terms I'm getting
$V=bc(c-b)-ac(c-a)+ab(b-a)$
But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.
 A: Subtract the first column from the second column and factor out a common term from the second column.
$\left|\begin{matrix}
1 & 1 & 1\\ 
a & b & c\\ 
a^2 & b^2 & c^2
\end{matrix}\right|=\left|\begin{matrix}
1 & 0 & 1\\ 
a & b-a & c\\ 
a^2 & b^2-a^2 & c^2
\end{matrix}\right|=(b-a)\left|\begin{matrix}
1 & 0 & 1\\ 
a & 1 & c\\ 
a^2 & b+a & c^2
\end{matrix}\right|$
Now do that again.
$(b-a)\left|\begin{matrix}
1 & 0 & 1\\ 
a & 1 & c\\ 
a^2 & b+a & c^2
\end{matrix}\right|=
(b-a)\left|\begin{matrix}
1 & 0 & 0\\ 
a & 1 & c-a\\ 
a^2 & b+a & c^2-a^2
\end{matrix}\right|=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\ 
a & 1 & 1\\ 
a^2 & b+a & a+c
\end{matrix}\right|$
Last, subtract the second column from the third column
$=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\ 
a & 1 & 0\\ 
a^2 & b+a & c-b
\end{matrix}\right|$
That final discriminant is obviously $c-b$, and you're done.
A: Note that
$$
V=bc(c-b)-ac(c-a)+ab(b-a) = - (a - b)(a - c)(b - c).
$$
A: $$\begin{align}
V &= bc^2 - cb^2 - ac^2 + ca^2 + ab^2 - ba^2\\
&=bc^2 \color{fuchsia}{- bac} - cb^2 + ab^2 - ac^2 + ca^2 \color{fuchsia}{+ abc} - ba^2\\
&=b(c^2-ac-bc+ab)-a(c^2-ac+bc+ab)\\
&=(b-a)(c^2-ac-bc+ab)\\
&=(b-a)(c-b)(c-a)
\end{align}$$
A: One way to prove this identity might use some properties of matrices:
$V=\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$
Multiply the first row by $a$ and subtract from the second row, anotate this result as the new second row. Use the original second row and multiply it by $a$ and subtract from the third row, anotate this result as the new third row.
$2R-1R\times a, 3R-2R\times a$
$\left|\begin{matrix}1&1&1\\0&b-a&c-a\\0&b^2-ab&c^2-ac\end{matrix}\right|$
Then using the known identity:
$1\left|\begin{matrix}b-a&c-a\\b^2-ab&c^2-ac\end{matrix}\right|-1\left|\begin{matrix}0&c-a\\0&c^2-ac\end{matrix}\right|+1\left|\begin{matrix}0&b-a\\0&b^2-ab\end{matrix}\right|$
This becomes into:
$V=1\left|\begin{matrix}b-a&c-a\\b^2-ab&c^2-ac\end{matrix}\right|$
$V=1\left|\begin{matrix}b-a&c-a\\b(b-a)&c(c-a)\end{matrix}\right|$
$V=1 (b-a)(c-a)\left|\begin{matrix}1&1\\b&c\end{matrix}\right|$
$V= (b-a)(c-a)(c-b)$
Thus that's it how it can proved. This works using the identity of factoring terms from the columns in the Vandermonde's matrix.
