How to solve this determinant 
Question Statment:-
  Show that
  \begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\ 
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\ 
\end{vmatrix}
=2abc(a+b+c)^3
\end{align*}


My Attempt:-
$$\begin{aligned}
&\begin{vmatrix}
\\(a+b)^2 & ca & bc \\ 
\\ca & (b+c)^2 & ab \\
\\bc & ab & (c+a)^2 \\\
\end{vmatrix}\\
=&\begin{vmatrix}
\\a^2+b^2+2ab & ca & bc \\ 
\\ca & b^2+c^2+2bc & ab \\
\\bc & ab & c^2+a^2+2ac \\\
\end{vmatrix}\\
=&\dfrac{1}{abc}\begin{vmatrix}
\\ca^2+cb^2+2abc & ca^2 & b^2c \\ 
\\ac^2 & ab^2+ac^2+2abc & ab^2 \\
\\bc^2 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow cC_1, C_2\rightarrow aC_2, C_3\rightarrow bC_3)\\\\
=&\dfrac{2}{abc}\times\begin{vmatrix}
\\ca^2+cb^2+abc & ca^2 & b^2c \\ 
\\ab^2+ac^2+abc & ab^2+ac^2+2abc & ab^2 \\
\\bc^2+a^2b+abc & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\\\\
&\qquad (C_1\rightarrow C_1+C_2+C_3)\\\\
=&\dfrac{2abc}{abc}\left(\begin{vmatrix}
\\a^2+b^2 & a^2 & b^2 \\ 
\\b^2+c^2 & b^2+c^2+2bc & b^2 \\
\\c^2+a^2 & a^2 & c^2+a^2+2ac \\\
\end{vmatrix}+
\begin{vmatrix}
\\1 & ca^2 & b^2c \\ 
\\1 & ab^2+ac^2+2abc & ab^2 \\
\\1 & a^2b & bc^2+a^2b+2abc \\\
\end{vmatrix}\right)
\end{aligned}$$
The second determinant in the last step can be simplified to 
\begin{vmatrix}
\\1 & ca^2 & b^2c \\ 
\\0 & ab^2+ac^2+2abc-ca^2 & ab^2-b^2c \\
\\0 & a^2b-ca^2 & bc^2+a^2b+2abc-b^2c \\\
\end{vmatrix}
I couldn't proceed further with this, so your help will be appreciated and if any other simpler way is possible please do post it too.
 A: One can use factor theorem to get a simpler solution. If we put $a=0$, we get 
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\ 
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\ 
\end{vmatrix}
=\begin{vmatrix}
b^2 & 0 & bc \\ 
0 & (b+c)^2 & 0 \\
bc & 0 & c^2 \\ 
\end{vmatrix} = 0
\end{align*}
Hence $a$ is a factor. Similarly $b, c$ are factors. Again, put $a+b+c=0$, we get
\begin{align*}
\begin{vmatrix}
c^2 & ca & bc \\ 
ca & a^2 & ab \\
bc & ab & b^2 \\ 
\end{vmatrix}
=abc\begin{vmatrix}
c & a & b \\ 
c & a & b \\
c& a & b \\ 
\end{vmatrix} 
\end{align*}
Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$. Putting $a=b=c=1$, we obtain 
\begin{align*}
27k = \begin{vmatrix}
4 & 1 & 1 \\ 
1 & 4 & 1 \\
1 & 1 & 4 \\ 
\end{vmatrix} =54
\end{align*}
and $k=2$. Thus the given determinant equals $2abc(a+b+c)^3$
A: Let me try. You have $$LHS = [(a+b)(b+c)(c+a)]^2 + 2(abc)^2 - \sum b^2c^2(b+c)^2.$$
Note that $(a+b)(b+c)(c+a) = \sum bc(b+c) + 2abc$.
Then, $$LHS = \left(\sum bc(b+c)\right)^2 + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right) - \sum b^2c^2(b+c)^2$$
$$=2\sum a^2bc(a+b)(a+c) + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right)$$
$$ =2abc\left(\sum a(a+b)(a+c) + 3abc + 2\sum bc(b+c)\right) $$
$$ = 2abc \left(\sum a^3 + 6abc + 3\sum bc(b+c)\right) $$
$$ = 2abc \left(a+b+c\right)^3 $$
