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?

Try this. Multiply column $1$ by $c$, column $2$ by $a$ and column $3$ by $b$. Then
\begin{aligned}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
&=\frac{1}{abc}\det\begin{pmatrix}
a^2c & b^2a & c^2b\\
abc & abc & abc\\
b^2c& c^2a & a^2b
\end{pmatrix}\\
&=\det\begin{pmatrix}
a^2c & b^2a & c^2b\\
1& 1 & 1\\
b^2c& c^2a& a^2b
\end{pmatrix}\\
&=\det\begin{pmatrix}
a^2c-c^2b & b^2a-c^2b & c^2b\\
0& 0 & 1\\
b^2c-a^2b& c^2a-a^2b & a^2b
\end{pmatrix}\\
&=(a^2-bc)(c^2-ab)\det\begin{pmatrix}
c & -b & c^2b\\
0& 0 & 1\\
-b& a & a^2b
\end{pmatrix}\\
&=(a^2-bc)(c^2-ab)(b^2-ac).
\end{aligned}
(https://i.stack.imgur.com/D6GF9.jpg)
