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.