Using properties of determinants prove that: $\left|\begin{array}{ccc} (b+c)^{2} & a b & c a \\ a b & (a+c)^{2} & b c \\ a c & b c & (a+b)^{2} \end{array}\right|=2 a b c(a+b+c)^{3}$

Solution:

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