How to prove this equality of the determinant of matrix? 
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}


My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
a(b-a) & b(c-b) & c(a-c) \\
(b+a)(b-a) & (c+b)(c-b) & (a+c)(a-c)
\end{pmatrix}
\end{equation*}
But I think my direction is incorrect. Can anyone give me some hints or the solution of this question?

Hint:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} & bc^{-1} & ca^{-1}\\
1 & 1 & 1\\
ba^{-1}& cb^{-1} & ac^{-1}
\end{pmatrix}
\end{equation*}
\begin{equation*}
= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} -ca^{-1} & bc^{-1} -ca^{-1} & ca^{-1}\\
0 & 0 & 1\\
ba^{-1} -ac^{-1} & cb^{-1} -ac^{-1} & ac^{-1}
\end{pmatrix} 
\end{equation*}
$$ = -(ab)(bc)(ca)\left[ (ab^{-1} -ca^{-1} ) (cb^{-1} -ac^{-1}) -( ba^{-1} -ac^{-1})(bc^{-1} -ca^{-1})\right]  $$
$$  = -(ab)(bc)(ca) \mathbf{(a^2-bc)}\left[ (ab)^{-1}(cb^{-1} -ac^{-1}) + (ac)^{-1}(bc^{-1} -ca^{-1})\right] $$
