# Prove that:

Question:

Prove that $\left|\begin{array}{ccc}\frac{a^{2}+b^{2}}{c} & c & c \\ a & \frac{b^{2}+c^{2}}{a} & a \\ b & b & \frac{c^{2}+a^{2}}{b}\end{array}\right|=4 a b c$

Solution:

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

$=\frac{1}{a b c}\left|\begin{array}{ccc}a^{2}+b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2}\end{array}\right|$         [Multiplying $R_{1}, R_{2}$ and $R_{3}$ by $c, a$ and $b$ and then dividing by $a b c$ ]

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

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

$=\frac{1}{a b c}\left[-a^{2}\left|\begin{array}{cc}-2 b^{2} & -2 a^{2} \\ 0 & c^{2}+a^{2}-b^{2}\end{array}\right|+b^{2}\left|\begin{array}{cc}-2 b^{2} & -2 a^{2} \\ b^{2}+c^{2}-a^{2} & 0\end{array}\right| \quad\right.$ [Expanding along $C_{1}$ ]

$=\frac{1}{a b c}\left[-a^{2}\left\{-2 b^{2}\left(c^{2}+a^{2}-b^{2}\right)\right\}+b^{2}\left\{0+2 a^{2}\left(b^{2}+c^{2}-a^{2}\right)\right\}\right]$

$=\frac{1}{a b c}\left[-a^{2}\left\{-2 b^{2} c^{2}-2 b^{2} a^{2}+2 b^{4}\right\}+b^{2}\left\{2 a^{2} b^{2}+2 a^{2} c^{2}-2 a^{4}\right\}\right]$

$=\frac{1}{a b c}\left[2 a^{2} b^{2} c^{2}+2 a^{4} b^{2}-2 a^{2} b^{4}+2 a^{2} b^{4}+2 a^{2} b^{2} c^{2}-2 a^{4} b^{2}\right]$

$=\frac{1}{a b c} 4 a^{2} b^{2} c^{2}=4 a b c$

Hence proved.