# Solve this

Question:

$\left|\begin{array}{ccc}-a\left(b^{2}+c^{2}-a^{2}\right) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & -b\left(c^{2}+a^{2}-b^{2}\right) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & -c\left(a^{2}+b^{2}-c^{2}\right)\end{array}\right|=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}$

Solution:

$\Delta=\left|\begin{array}{ccc}-a\left(b^{2}+c^{2}-a^{2}\right) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & -b\left(c^{2}+a^{2}-b^{2}\right) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & -c\left(a^{2}+b^{2}-c^{2}\right)\end{array}\right|$

$=a b c\left|\begin{array}{ccc}-b^{2}-c^{2}+a^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & -c^{2}-a^{2}+b^{2} & 2 c^{2} \\ 2 a^{2} & 2 b^{2} & -a^{2}-b^{2}+c^{2}\end{array}\right| \quad$ [Taking out $a, b$ and $c$ common from $C_{1}, C_{2}$ and $C_{3}$ ]

$=a b c\left|\begin{array}{ccc}a^{2}+b^{2}+c^{2} & 2 b^{2} & 2 c^{2} \\ a^{2}+b^{2}+c^{2} & -c^{2}-a^{2}+b^{2} & 2 c^{2} \\ a^{2}+b^{2}+c^{2} & 2 b^{2} & -a^{2}-b^{2}+c^{2}\end{array}\right| \quad$ [Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ]

$=a b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}1 & 2 b^{2} & 2 c^{2} \\ 1 & -c^{2}-a^{2}+b^{2} & 2 c^{2} \\ 1 & 2 b^{2} & -a^{2}-b^{2}+c^{2}\end{array}\right|$             [Taking out $a^{2}+b^{2}+c$ common from $\left.C_{1}\right]$

$=a b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc}1 & 2 b^{2} & 2 c^{2} \\ 0 & -c^{2}-a^{2}-b^{2} & 0 \\ 0 & 0 & -a^{2}-b^{2}-c^{2}\end{array}\right|$     [Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}$ [Expanding]Hence proved.