$$\begin{vmatrix}1&a&a^2\\ 1&b&b^2\\ 1&c&c^2\end{vmatrix}=(b-a)(c-a)\begin{vmatrix}1&a&0\\ 0&1&b\\ 0&1&c\end{vmatrix}$$
I have been trying to solve this equation for about 2 weeks now but without any luck can any one help me with a simple solution for this? (if available ofcourse) or even a complex one doesn't really matter I just want to have a solution for it..
On
$$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}=\begin{vmatrix}1&a&a^2\\0&(b-a)&(b^2-a^2)\\0&(c-a)&(c^2-a^2)\end{vmatrix}=(b-a)(c-a)\begin{vmatrix}1&a&a^2\\0&1&(b+a)\\0&1&(c+a)\end{vmatrix} \dots (1)=(b-a)(c-a)\begin{vmatrix}1&a&a^2\\0&1&(b+a)\\0&0&(c-b)\end{vmatrix}=(b-a)(c-a)(c-b)\begin{vmatrix}1&a&a^2\\0&1&(b+a)\\0&0&1\end{vmatrix}=(b-a)(c-a)(c-b)\begin{vmatrix}1&a&0\\0&1&0\\0&0&1\end{vmatrix}=(b-a)(c-a)(c-b)\begin{vmatrix}1&0&0\\0&1&0\\0&0&1\end{vmatrix}=(a-b)(b-c)(c-a).$$
After $(1)$ do the following, $$\implies (b-a)(c-a)\begin{vmatrix}1&a&0\\0&1&b\\0&1&c\end{vmatrix} \text{ doing column operation.}$$
Hint: How is $$\begin{vmatrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{vmatrix}$$ related to $$\begin{vmatrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{vmatrix}?$$
How is $$\begin{vmatrix}1&a&a^2\\0&1&a+b\\0&1&a+c\end{vmatrix}$$ related to $$\begin{vmatrix}1&a&0\\0&1&b\\0&1&c\end{vmatrix}?$$