Using properties of determinants prove that $ \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} = \left(a+b+c\right) \left(a^2 + b^2 + c^2\right) $
$ \begin{aligned} \text{Let } \Delta & = \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow aC_1 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2 & b-c & c+b \\ a^2 + ac & b & c-a \\ a^2 - ab & b+a & c \end{vmatrix} \\ & \\ C_1 \rightarrow C_1 + bC_2 + cC_3 \implies \Delta & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2-bc+c^2+bc & b-c & c+b \\ a^2+ac+b^2+c^2-ac & b & c-a \\ a^2-ab+b^2+ab+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{1}{a} \begin{vmatrix} a^2+b^2+c^2 & b-c & c+b \\ a^2+b^2+c^2 & b & c-a \\ a^2+b^2+c^2 & b+a & c \end{vmatrix} \\ & \\ & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_2 \rightarrow R_2-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 1 & b+a & c \end{vmatrix} \\ & \\ R_3 \rightarrow R_3-R_1 \implies \Delta & = \dfrac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a+c & -b \end{vmatrix} \\ & \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left[-bc+\left(a+b\right)\left(a+c\right)\right]}{a} \left[\text{Note: Expanding along }C_1\right] \\ & = \dfrac{\left(a^2+b^2+c^2\right)\left(a^2+ab+ac+bc-bc\right)}{a} \\ & = \dfrac{\left(a^2+b^2+c^2\right) \left(a^2+ab+ac\right)}{a} \\ & = \left(a^2+b^2+c^2\right) \left(a+b+c\right) \end{aligned} $