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?

It has to be a degree-$6$ homogeneous polynomial in $a,\,b,\,c$ that vanishes if $a^2=bc$, because if $a/b=b/c=k$ the determinant is $c^6$ times$$\left|\begin{array}{ccc} k^{4} & k^{2} & 1\\ k^{3} & k & k^{2}\\ k^{2} & 1 & k^{4} \end{array}\right|=k^{4}\left(k^{5}-k^{2}\right)-k^{2}\left(k^{7}-k^{4}\right)=0.$$Repeating this logic for $2$ other factors, the determinant must be proportional to $(a^2-bc)(b^2-ca)(c^2-ab)$. The case $a=1,\,b=2,\,c=3$ lets you verify no proportionality constant is needed.
