Using properties of determinants prove that: $\left|\begin{array}{ccc} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$

Solution:

Operating $\mathrm{C}_{1} \rightarrow \mathrm{aCl}_{1}$
$\frac{1}{a}\left|\begin{array}{ccc} a^{2} & b-c & c+b \\ a^{2}+a c & b & c-a \\ a^{2}-a b & a+b & c \end{array}\right|$
Operating $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{b} \mathrm{C}_{2}+\mathrm{cC}_{3}$
$\begin{array}{l} =\frac{1}{a}\left|\begin{array}{ccc} a^{2}+b^{2}-b c+c^{2}+b c & b-c & c+b \\ a^{2}+a c+b^{2}+c^{2}-a c & b & c-a \\ a^{2}-a b+a b+b^{2}+c^{2} & a+b & c \end{array}\right| \\ =\frac{1}{a}\left|\begin{array}{lll} a^{2}+b^{2}+c^{2} & b-c & c+b \\ a^{2}+b^{2}+c^{2} & b & c-a \\ a^{2}+b^{2}+c^{2} & a+b & c \end{array}\right| \end{array}$
Taking $\left(a^{2}+b^{2}+c^{2}\right)$ common from $c_{1}$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & a+b & c \end{array}\right|$
Operating $R_{1} \rightarrow R_{1}-R_{3}, R_{2} \rightarrow R_{2}-R_{3}$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 0 & -c-a & b \\ 0 & -a & -a \\ 1 & a+b & c \end{array}\right|$
Operating $C_{2} \rightarrow C_{2}-C_{3}$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 0 & -(a+b+c) & b \\ 0 & 0 & -a \\ 1 & (a+b+c) & c \end{array}\right|$
Taking $(a+b+c)$ common from $c_{2}$
$=\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)\left|\begin{array}{ccc} 0 & -1 & b \\ 0 & 0 & -a \\ 1 & 1 & c \end{array}\right|$
Expanding with respect to $\mathrm{C}_{1}$
$\begin{array}{l} =\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)(a+b+c) \times 1 \times(0-(-a)) \\ =\frac{1}{a}\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)(a) \\ =\left(a^{2}+b^{2}+c^{2}\right)(a+b+c) \end{array}$