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.

Solution 1:

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 $2abc(a+b+c)^3$

Solution 2:

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, $$LHS = \left(\sum bc(b+c)\right)^2 + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right) - \sum b^2c^2(b+c)^2$$

$$=2\sum a^2bc(a+b)(a+c) + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right)$$

$$ =2abc\left(\sum a(a+b)(a+c) + 3abc + 2\sum bc(b+c)\right) $$

$$ = 2abc \left(\sum a^3 + 6abc + 3\sum bc(b+c)\right) $$

$$ = 2abc \left(a+b+c\right)^3 $$