<s>[INST] $\begin{vmatrix} 1 & a &bc \\ 1& b & ac\\ 1&c & ab \end{vmatrix}=\begin{vmatrix} 1 & a &a^2 \\ 1& b&b^2 \\ 1& b & c^2 \end{vmatrix}$ 
Prove that  \begin{align}\begin{vmatrix}
1 & a &bc \\ 
 1& b & ac\\ 
 1&c  & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\ 
 1&  b&b^2 \\ 
 1& b & c^2
\end{vmatrix}\\&=(c-a)(b-a)(c-b)\begin{vmatrix}
1 & a & a^2\\ 
 0& 1 &b+a \\ 
 0& 0 & 1
\end{vmatrix}\\\\
&=(c-a)(b-a)(c-b)\end{align}

I got the last two equalities but I did't get first two. I need help with first two
 [/INST] \begin{align}\begin{vmatrix}
1 & a &bc \\ 
 1& b & ac\\ 
 1&c  & ab
\end{vmatrix}&=\begin{vmatrix}
1 & a &a^2 \\ 
 1&  b&b^2 \\ 
 1& c & c^2
\end{vmatrix}\\\\
\begin{vmatrix}
1 & a &bc \\ 
 0& b-a & ac-bc\\ 
 0&c-a  & ab-bc
\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 &bc \\ 
 0& 1& -c\\ 
 0&1  &-b
\end{vmatrix}&=(b-a)(c-a)\begin{vmatrix}
1 & a &a^2 \\ 
 0& 1&b+a \\ 
 0& 1 & c+a
\end{vmatrix}\\\\
(b-a)(c-a)\begin{vmatrix}
1 & a &bc \\ 
 0& 1& -c\\ 
 0&0  &c-b
\end{vmatrix}&=(b-a)(c-a)\begin{vmatrix}
1 & a &a^2 \\ 
 0& 1&b+a \\ 
 0& 0& c-b
\end{vmatrix}\end{align}
</s>