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?
 A: 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.
A: 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)
A: 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*}
$$ = -(ab)(bc)(ca)\left[ (ab^{-1} -ca^{-1} ) (cb^{-1} -ac^{-1}) -( ba^{-1} -ac^{-1})(bc^{-1} -ca^{-1})\right]  $$
$$  = -(ab)(bc)(ca) \mathbf{(a^2-bc)}\left[ (ab)^{-1}(cb^{-1} -ac^{-1}) + (ac)^{-1}(bc^{-1} -ca^{-1})\right] $$
A: My steps to calculate your determinant. With $R$ I indicate the row and with $C$ the column.
$$C_2=C_2-\left(\frac{b^{2}}{a^{2}}\right)C_1 $$
$$\left| \begin{array}{ccc} a^{2} & 0 & c^{2} \\\\ a b & b c - \frac{b^{3}}{a} & a c \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & a^{2} \end{array} \right| \tag{first step}$$
Subtract column $1$ multiplied by $c^2/a^2$ from column $3$; you will have
$$\left| \begin{array}{ccc} a^{2} & 0 & 0 \\\\ a b & b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right| \tag{second step}$$
Using the first rule of Laplace:
$$\left| \begin{array}{ccc} a^{2} & 0 & 0 \\\\ a b & b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|=\left(a^{2}\right) \cdot(-1)^{1+1}\cdot \left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|+0 \cdot(-1)^{1+2}\cdot \left| \begin{array}{cc} a b & a c - \frac{b c^{2}}{a} \\\\ b^{2} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|+0 \cdot(-1)^{1+3}\cdot \left| \begin{array}{cc} a b & b c - \frac{b^{3}}{a} \\\\ b^{2} & c^{2} - \frac{b^{4}}{a^{2}} \end{array} \right|=a^{2} \left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|$$
Considering the last determinant you have:
$$\left| \begin{array}{cc} b c - \frac{b^{3}}{a} & a c - \frac{b c^{2}}{a} \\\\ c^{2} - \frac{b^{4}}{a^{2}} & \frac{a^{4} - b^{2} c^{2}}{a^{2}} \end{array} \right|=\left(b c - \frac{b^{3}}{a}\right)\cdot\left(\frac{a^{4} - b^{2} c^{2}}{a^{2}}\right)-\left(a c - \frac{b c^{2}}{a}\right)\cdot\left(c^{2} - \frac{b^{4}}{a^{2}}\right)=\frac{\left(a^{2} - b c\right) \left(a b - c^{2}\right) \left(a c - b^{2}\right)}{a^{2}}$$
Hence:
$$\left| \begin{array}{ccc} a^{2} & b^{2} & c^{2} \\\\ a b & b c & a c \\\\ b^{2} & c^{2} & a^{2} \end{array} \right|=a^{2} \cdot \left(\frac{\left(a^{2} - b c\right) \left(a b - c^{2}\right) \left(a c - b^{2}\right)}{a^{2}}\right)=(a^2-bc)(b^2-ca)(c^2-ab)$$
