Question

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

Answer 1

$$\begin{gathered} ∆=\left|\begin{array}{ccc}-a(b^2+c^2-a^2)& 2b^3& 2c^3\\ 2a^3& -b(c^2+a^2-b^2)& 2c^3\\ 2a^3& 2b^3& -c(a^2+b^2-c^2)\end{array}\right| \\ =abc\left|\begin{array}{ccc}-b^2-c^2+a^2& 2b^2& 2c^2\\ 2a^2& -c^2-a^2+b^2& 2c^2\\ 2a^2& 2b^2& -a^2-b^2+c^2\end{array}\right|\left[\mathrm{Taking}\mathrm{out}a,b\mathrm{and}c\mathrm{common}\mathrm{from}{C}_1,C_2\mathrm{and}{C}_3\right] \\ =abc\left|\begin{array}{ccc}{a}^2+b^2+c^2& 2b^2& 2c^2\\ a^2+b^2+c^2& -c^2-a^2+b^2& 2c^2\\ a^2+b^2+c^2& 2b^2& -a^2-b^2+c^2\end{array}\right|\left[\mathrm{Applying}{C}_1\to C_1+C_2+C_3\right] \\ =abc(a^2+b^2+c^2)\left|\begin{array}{ccc}1& 2b^2& 2c^2\\ 1& -c^2-a^2+b^2& 2c^2\\ 1& 2b^2& -a^2-b^2+c^2\end{array}\right|\left[\mathrm{Taking}\mathrm{out}{a}^2+b^2+c\mathrm{common}\mathrm{from}{C}_1\right] \\ =abc(a^2+b^2+c^2)\left|\begin{array}{ccc}1& 2b^2& 2c^2\\ 0& -c^2-a^2-b^2& 0\\ 0& 0& -a^2-b^2-c^2\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_1\mathrm{and}{R}_3\to R_3-R_1\right] \\ =abc(a^2+b^2+c^2{)}^3\left[\mathrm{Expanding}\right] \end{gathered}$$

Hence proved.