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# Using the properties of determinants, prove the following : $\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & c+a \\ a+b & b+c & c+a \end{vmatrix} = 2(a+b+c)(ab+bc+ca-a^2-b^2-c^2).$

Toolbox:
• Elementary transformation can be done by interchanging any two rows or any two columns.
• The addition to the elements of any row or column the corresponding elements of any other row or column multiplied by any non-zero number.
Given:

$\Delta=\begin{vmatrix}b+c & c+a & a+b\\c+a & a+b & b+c\\a+b & b+c & c+a\end{vmatrix}$

Apply $R_1\rightarrow R_1+R_2+R_3$

$\Delta=\begin{vmatrix}2a+2b+2c & 2a+2b+2c & 2a+2b+2c\\c+a & a+b & b+c\\a+b & b+c & c+a\end{vmatrix}$

Taking 2(a+b+c) as a common factor from $R_1$

$\Delta=2(a+b+c)\begin{vmatrix}1 & 1 & 1\\c+a & a+b & b+c\\a+b & b+c & c+a\end{vmatrix}$

Apply $C_1\rightarrow C_1-C_2$ and $C_2\rightarrow C_2-C_3$

$\Delta=2(a+b+c)\begin{vmatrix}0 & 0 & 1\\c-b & a-c & b+c\\a-c & b-a & c+a\end{vmatrix}$

Now expanding along $R_1$

$\;\;\;=2(a+b+c)[0-0+1[(c-b)(b-a)-(a-c)(a-c)]$

$\;\;\;=2(a+b+c)[bc-ac-b^2+ab-a^2+ac+ac-c^2]$

$\;\;\;=2(a+b+c)[ab+bc+ac-a^2-b^2-c^2]$

Hence proved.