Expressing a $3\times 3$ determinant as the product of four factors

Question

I am attempting to express the determinant below as a product of four linear factors $$\begin{vmatrix} a & bc & b+c\\ b & ca & c+a\\ c & ab & a+b\\ \end{vmatrix} = a\begin{vmatrix} ca & c+a\\ ab & a+b\\ \end{vmatrix} - bc\begin{vmatrix} b & c+a\\ c & a+b\\ \end{vmatrix} +(b+c)\begin{vmatrix} b & ca\\ c & ab\\ \end{vmatrix}$$ This is as far as I get before it gets too messy $$ =a^3(c-b)-bc\{(b-c)(b+c)+a(b-c)\}+a(b-c)(b+c)^2 $$ But I cant seem to arrive at the answer in the book, which is given as $$ (a-b)(b-c)(c-a)(a+b+c) $$

Am I doing something wrong as I have been stuck on this question for three days.

Thanks in advance! $$ $$

Answer 1

Start by adding the 1st to the 3rd column to create a column of $a+b+c$'s. Then subtract 3rd row from 2nd & 1st ones to make two out of three entries in that column zero. Now expand wrt that column.

Answer 2

Steps:

1. Add first column to third column & take (a+b+c) out from third column.
2. Now Subtract Row2 from Row1 & take a-b out from Row1
3. Subtract Row3 from Row2 & take b-c out from Row2.
4. Now Multiply -1 to Row1 & Add Row1 & Row2.
5. Now take c-a out from Row1.
6. Now calculate determinant as usual.

Answer 3

$$\begin{vmatrix} a & bc & b+c\\ b & ca & c+a\\ c & ab & a+b \end{vmatrix} = \begin{vmatrix} a & bc & b+c\\ b -a & c(a-b)& a-b\\ c -a& b(a-c)& a-c \end{vmatrix}=(a-b)(c-a) \begin{vmatrix} a & bc & b+c\\ -1&c&1\\ 1&-b&-1 \end{vmatrix}$$ $$=(a-b)(c-a)\begin{vmatrix} a& bc& b+c\\ -1&c&-1\\ 0&b-c&0 \end{vmatrix} =(a-b)(c-a)(b-c)(a+b+c)$$