Difficulty with factorizing determinant

Question

I came across this determinant and its just proved difficult to express as a product of in linear factors $$\begin{vmatrix} 1 & 1 & 1 \\a^2 & b^2 & c^2 \\(b+c)^2 & (c+a)^2 & (a+b)^2 \end{vmatrix}$$ I have tried the basic operation combining rows and column but I can't just get to the answer $2(a-b)(b-c)(a+b+c)(c-a)$. I really need a guide or hints to proceed.

Answer 1

Let us call $\Delta$ this determinant. Do the following operation: replace the last column by its difference with the second one ($C_3\leftarrow C_3-C_2$) to get $$ \Delta=\begin{vmatrix} 1 & 1 &0\\a^2 & b^2 & c^2-b^2 \\(b+c)^2 & (c+a)^2 & (a+b)^2-(c+a)^2 \end{vmatrix}. $$ Since $(a+b)^2-(c+a)^2=\left(b-c\right)(2a+b+c)$, we can use multilinearity of the determinant to get $$ \Delta=(b-c)\begin{vmatrix} 1 & 1 &0\\a^2 & b^2 & -(b+c) \\(b+c)^2 & (c+a)^2 & 2a+b+c \end{vmatrix}. $$ Then We do $C_2\leftarrow C_2-C_1$ to get $$ \Delta=(b-c)(a-b)\begin{vmatrix} 1 & 0&0\\a^2 & -(a+b) & -(b+c) \\(a+c)^2 & a+b+2c & 2a+b+c \end{vmatrix}. $$ Expanding with respect to the first line gives $$ \Delta=(b-c)(a-b)\begin{vmatrix} -(a+b) & -(b+c) \\ a+b+2c & 2a+b+c \end{vmatrix}. $$ Doing $L_2\leftarrow L_2+L_1$ gives $$ \Delta=(b-c)(a-b)\begin{vmatrix} -(a+b) & -(b+c) \\ 2c & 2a \end{vmatrix} =2(b-c)(b-a)\begin{vmatrix} a+b & b+c \\ c & a \end{vmatrix}. $$ Finally, do $C_2\leftarrow C_2-C_1$ to get the wanted result.