LA2.1.18

Show that $\begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}=3abc-a^3-b^3-c^3\,$

Applying $C_1+C_3,C_2-C_3\,$ we get

$\begin{vmatrix} a+b+c & -b & a \\ a+b+c & -c & b \\ a+b+c & -a & c \end{vmatrix}\,$

$(a+b+c)(-1)\begin{vmatrix} 1 & b & a \\ 1 & c & b \\ 1 & a & c \end{vmatrix}\,$

Applying $R_2-R_1,C_2-C_3\,$ we have

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

$-(a+b+c)[(c-b)(c-a)-(a-b)(b-a)]=-(a+b+c)[a^2+b^2+c^2-ab-bc-ca]=-(a^3+b^3+c^3-3abc)=3abc-a^3-b^3-c^3\,$ =RHS