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

Solution:

Taking $a, b, c$ from $C_{1}, C_{2}, C_{3}$
$=a b c\left|\begin{array}{ccc} -b^{2}-c^{2}+a^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & b^{2}-c^{2}-a^{2} & 2 c^{2} \\ 2 a^{2} & 2 b^{2} & \left.-a^{2}-b^{2}+c^{2}\right) \end{array}\right|$
Operating $R_{1} \rightarrow R_{1}-R_{3}, R_{2} \rightarrow R_{2}-R_{3}$
$=a b c\left|\begin{array}{ccc} -b^{2}-c^{2}-a^{2} & 0 & a^{2}+b^{2}+c^{2} \\ 0 & -\left(a^{2}+b^{2}+c^{2}\right) & a^{2}+b^{2}+c^{2} \\ 2 a^{2} & 2 b^{2} & -a^{2}-b^{2}+c^{2} \end{array}\right|$
Taking $\left(a^{2}+b^{2}+c^{2}\right)$ common from $R_{1}, R_{2}$
$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{2}\left|\begin{array}{ccc} -1 & 0 & 1 \\ 0 & -1 & 1 \\ 2 a^{2} & 2 b^{2} & -a^{2}-b^{2}+c^{2} \end{array}\right|$
Operating $R_{3} \rightarrow R_{3}+R_{1}+R_{2}$
$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{2}\left|\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 2 a^{2} & 2 b^{2} & a^{2}+b^{2}+c^{2} \end{array}\right|$
Taking $\left(a^{2}+b^{2}+c^{2}\right)$ common from $C_{3}$
$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}\left|\begin{array}{ccc} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 2 a^{2} & 2 b^{2} & 1 \end{array}\right|$
Expanding with $C_{3}$
$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3} \times 1 \times(1-0)$
$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}$
Hence proved